Dual Nature Of Radiation And Matter
Class 12th Physics Part Ii CBSE Solution
Exercises
Question 1.Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Answer:
(a) X-ray produced by the electron therefore has maximum photon energy (Ex) equal to energy (E) of the electron.
E = eV …(1)
Where magnitude of charge of the electron ‘e’ = 1.6 × 10-19 C
Accelerating potential is V = 30 × 103 V 3
Putting in equation (1) we get,
⇒ E = 1.6 × 10-19 C × 30 × 103 J
Also, Ex = hν
Where, Ex = maximum energy of the X-ray
And h = plank’s constant = 6.63 × 10-34 J. sec
ν = max. frequency of X-ray
⇒ vmax = Emax/h
⇒ 
⇒ vmax= 7.24 × 1018 Hz
Max. frequency of X-ray is 7.24 × 1018 Hz
(b) Minimum wave length (λ) of X-ray is given by the following relation.
Ex = hν = hc/λ
∴ λ = hc / Ex
c = speed of light = 3 × 108 m/s
⇒ 
⇒ λ = 0.041nm
Minimum wave length of X-ray is equal to 0.041nm
Question 2.
The work function of caesium metal is 2.14 eV. When the light of frequency 6 × 1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons,
(b) Stopping potential, and
(c) maximum speed of the emitted photoelectrons?
Answer:
(a) maximum energy (K. Emax) of emitted electron is given by the following equation
K. Emax = hν – W0 ……… equation no. 1
Where h = plank’s constant = 6.63 × 10-34 J. sec
ν = max. frequency of X-ray
W0 = work function (minimum energy required to emit an electron from neutral atom)
W0 = 2.14 eV
And, ν = 6 × 1014 Hz
From equation No. 1
⇒ 
⇒ KEmax = 0.34 eV
Or in Joules,
KEmax = 0.34 × 1.6 × 10-19 = 0.546 × 10-19 J
maximum KE of emitted electron is 0.546 × 10-19 J
(b) Stopping Potential is calculated by following relation
eV0 = K.Emax
Magnitude of charge of the electron ‘e’ = 1.6 × 10-19 C
V0 = Stopping Potential
⇒ 
⇒ V0 = 0.34 V
Stopping Potential of emitted electron is 0.34 V
(c) K.Emax = 1/2 mv2max
m = mass of the electron = 9.1 × 10-31 kg
vmax = maximum velocity of electron
v2max = 2 K.Emax / m
vmax = (2 × 0.546 × 10-19 J / 9.1 × 10-31)1/2
= 347 km /sec
Maximum velocity of emitted electron is 347 km / sec.
Question 3.
The photoelectric cut-off voltage in a certain experiment is 1.5 V.
What is the maximum kinetic energy of photoelectrons emitted?
Answer:
The maximum kinetic energy of the emitted electron is equal to work done on the electron to stop it to reach to the anode. Therefore
K.Emax = 1/2 mv2max = eV0
Magnitude of charge of the electron ‘e’ = 1.6 × 10-19 C
V0 = Stopping Potential
K.Emax = maximum kinetic energy of emitted electron
m = mass of the electron = 9.1 × 10-31 kg
vmax = maximum velocity of electron
K.Emax = eV0 = 1.6 × 10-19 × 1.5 = 1.5 Ev
= 2.4 × 10-19 J
maximum KE of electron is 2.4 × 10-19 J
Question 4.
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Find the energy and momentum of each photon in the light beam,
Answer:
(a) Energy of photon E = hv = hc/λ
Where h = plank’s constant = 6.63 × 10-34 J. sec
ν = max. frequency of X-ray
c = speed of light = 3 × 108 m/s
λ = minimum wavelength of X-ray
E = 6.63 × 10-34 × 3 × 108 / 632.8 x10-9
∴ E = 3.14 × 10-19 J
Now, p = h / λ
p = momentum of photon
p = 6.63 × 10-34/632.8 x10-9
∴ p = 1.05 × 10-27 kg-m/sec
Maximum energy and momentum of the electron are 3.14 × 10-19 J and 1.05 × 10-27 kg-m/sec respectively.
(b) No. of photons ‘n’ per sec. is equal to total power radiated divided by number of photons (it is assumed that power of each and every photon is same and all the photons are falling on the target).
‘n’ = P/E
Where P = radiated power of LASSER
n = 9.42 × 106/3.14 × 10-19
n = 3 × 1016 photons/sec.
No. of photons falling per second is equal to 3 × 1016 photons/sec.
(c) mv = p
m = mass of Hydrogen = mass of proton = 1.67 × 10-27 kg
v = speed of Hydrogen
p = momentum of photon as well as Hydrogen atom
v = p/m
v = 1.05 × 10-27/1.67 × 10-27 = 0.63 m/sec.
Required speed of Hydrogen atom is 0.63 m/sec.
Question 5.
The energy flux of sunlight reaching the surface of the earth is 1.388 × 103 W/m2. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Answer:
Energy of a photon E = hv = hc / λ
Where h = plank’s constant = 6.63 × 10-34 J. sec
ν = average frequency of sunlight
c = speed of light = 3 × 108 m/s
λ = average wavelength of sunlight
E = 6.63 × 10-34 × 3 × 108 / 550 × 10-9 = 3.61 × 10-19 J
ϕ = energy flux per unit area (sq. meter) per sec. falling on the earth.
n = let no. of photons falling per sec. per sq. meter on earth surface.
ϕ = n × E
n = ϕ/E
n = 1.388 × 103/3.61 × 10-19 = 3.85 × 1021 photons/m2.sec
no. of photons falling on Earth surface per square meter per sec. is 3.85 × 1021 photons/m2.sec.
Question 6.
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10–15 V s.
Calculate the value of Planck’s constant.
Answer:
Photoelectric equation
eV0 = hv - ϕ0 ………… eq.01
Where e = Charge on an electron = 1.6 × 10-19 C
V0 = Stopping Potential
h = plank’s constant
v = frequency of photon
ϕ0 = Work function of the metal
Rewriting equation no. 1
V0 = hv/e - ϕ0/e ………………………………equ.02
h/e is slop of the line represented by above equation02.
Therefore h/e = slop of the line = 4.12 × 10-15 Vs
h = 1.6 × 10-19 × 4.12 × 10-15 = 6.594 × 10-34 Js
Value of plank’s constant ‘h’ from above experiment is equal to 6.594 × 10-34 Js.
Question 7.
A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?
Answer:
(a) Energy of photon E = hv = hc/λ
Where h = plank’s constant = 6.63 × 10-34 J. sec
ν = frequency of light
c = speed of light = 3 × 108 m/s
λ = wavelength of light
E = 6.63 × 10-34 × 3 x108 / 589 × 10-9
= 3.38 × 10-19 J
Energy of a photon is equal to 3.38 × 10-19 J.
(b) Rate of photon falling on the sphere = power of lamp/energy per photon
∴ Rate of photon = 100 / 3.38 × 10-19J = 3.0 × 1020 photons/sec.
Rate at which photons are falling on the sphere is equal to 3.0 × 1020 photons/sec.
Question 8.
The threshold frequency for a certain metal is 3.3 × 1014 Hz. If the light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.
Answer:
Photoelectric equation
eV0 = hv - ϕ0 ……………… equation(1)
Where e = Charge on an electron = 1.6 × 10-19 C
V0 = Stopping Potential
h = plank’s constant
v = frequency of photon
ϕ0 = Work function of the metal
Threshold frequency is the minimum frequency at which photoemission starts taking place from a metal surface. It is represented by v0.
At threshold frequency, stopping potential is zero and energy of the emitted electron is also zero. Energy provided by the incident photon is just sufficient to eject out electrons from the metal surface.
ϕ0 = hv0 ....................equ.02
V0 = Threshold frequency
Putting value of ϕ0 from equ.02 to equ.01
eV0 = hv – hv0
∴ V0 = h (v-v0)/ e = 6.63 × 10-34 × (8.2-3.3) × 1014/1.6 × 10-19
= 2.03 V
For photo emission cut of potential is equal to 2.03 V
Question 9.
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Answer:
For photoemission energy ‘E’ of incident radiation must be more than or equal to work function ‘ϕ0’ of the metal.
E 
ϕ0 (The Condition for photoelectric emission)
E = hv = hc/λ
Where h = plank’s constant = 6.63 × 10-34 J. sec
ν = frequency of radiation
c = speed of light = 3 × 108 m/s
λ = wavelength of radiation
E = 6.63 × 10-34 × 3 × 108 / 330 × 10-9
= 6.018 × 10-19 J = 6.018 × 10-19/1.6 × 10-19 eV
= 3.76 eV
ϕ0 = 4.2 eV as per question.
E < ϕ0, therefore photoemission will not take place.
No photoemission will take place as work function is more than the photon energy of the incident radiation.
Question 10.
Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Answer:
Photoelectric equation
K. Emax = hv – hv0 ……………………………………….equ.01
Where K.Emax = maximum kinetic energy of the photoelectron
h = plank’s constant
v = frequency of photon
v0 = threshold frequency
K. Emax = 1/2 mv2max ........................................equ.02
Where ‘m’ = mass of the electron = 9.1 × 10-31 kg
Vmax = maximum speed of photoelectrons
From equ.01 and equ.02
1/2 mv2max = hv - hv0
∴ v0 = v - 1/2 mv2max / h
⇒ 
⇒ v0 = 7.21 × 1014 –2.47 × 1014
⇒ v0 = 4.74 × 1014 Hz
Threshold frequency for photoemission is equal to 4.74 × 1014 Hz.
Question 11.
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
Answer:
In photoelectric effect an electromagnetic radiation (Can be Visible Light) is incident on a metal, if the energy of incident radiation is more than work function of the metal, then electrons are ejected out of the metal surface with some kinetic energy
Work function is the minimum Energy required to just pull out an electron from metal surface.
the setup for photoelectric effect has been shown in figure
Now the energy supplied to electron through photon (assuming light to be particle) is used up in overcoming the internal atomic forces to come to surface of metal (equal to work function of metal) and end extra energy is converted to kinetic energy of particle
So the maximum kinetic energy of an electron can be given by the photoelectric equation
Kmax = E – ϕo
Where Kmax is the maximum kinetic energy of electron.
E is the energy of incident radiation.
ϕo is the work function of metal.
We know energy of an electromagnetic radiation can be given by relation
E = h𝜈
Where h is Planck’s constant.
𝜈 is the frequency of radiation.
So, above equation becomes
Kmax = h𝜈 – ϕo
Now to stop the moving electrons a potential difference is such applied which applies force in opposite direction to the motion, if the energy transferred to electron due to potential difference in equal to maximum kinetic energy of electron then the electrons completely stop moving. We know energy transferred or work done by a potential difference on a charged particle is given by relation
W = qVo
Where W is the energy transferred by a potential difference of Vo volts to a charge particle having charge q
Now for electrons to completely stop we have,
Kmax = W
Or Kmax = qVo
Since particle is electron so we have
q = e
e is charge of electron, e = 1.6 × 10-19 C
i.e. Kmax = eVo
So, the photoelectric equation becomes
eVo = h𝜈 – ϕo
or we can say the work function is given as
ϕo = h𝜈 – eVo
where Vo is stopping Potential Difference
we know the relation for any electromagnetic wave we have
c = 𝜈𝜆
Where 𝜈 is the frequency and 𝜆 is wavelength of electromagnetic radiation having velocity C
So, re arranging we get
𝜈 = c/𝜆
So the photo electric equation becomes
ϕo = hc/𝜆 – eV
We know value of Planck’s constant
h = 6.6 × 10-34Js
charge of electron
e = 1.6 × 10-19 C
the speed of light
C = 3 × 108 m/s
light is incident from argon laser Light of wavelength
𝜆 = 488 nm = 488 × 10-9 m
We are given the stopping potential difference
Vo = 0.38 V
Putting thse values in equation to find work function of emitter ϕo
ϕo = hC/𝜆 – eVo
we have
We get
ϕo = 4.05 × 10-19J – 0.608 × 10-19 J = 3.46 × 10-19J
so work function of emitter metal is 3.46 × 10-19J
converting it to eV
1 eV = 1.6 × 10-19J
i.e.
1J = (1/1.6 × 10-19)eV
So we get work function of emitter metal as
We get ϕo = 2.16 eV
So the work function of emitter metal is 2.16 eV
Question 12.
Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Answer:
(a) When an charged particle having charge q is accelerated by a potential difference V volts it gain kinetic energy given by
K = Vq, which imparts it with velocity
We know kinetic energy of a Particle with mass m kg and velocity v is given by
Equating both equations
We get velocity of the particle as
But we know momentum is given by relation
P = mv
Where P is the momentum of particle having mass m and moving with velocity v
So we get
Here Potential difference
V = 56 volts
q = 1.6 × 10-19 C
m = 9.1 × 10-31Kg
so momentum of the electron,
So momentum of electron is 4.04 × 10-24 Kgms-1
(b) Now we know de Broglie wavelength of a Particle is given by relation
𝜆 = h/mv
Where 𝜆 is de Broglie wavelength of the particle having mass m and moving with velocity v
But we know momentum is given by relation
P = mv
Where P is the momentum of particle having mass m and moving with velocity v
So substituting we get de Broglie wavelength of Particle as
𝜆 = h/P
h = 6.6 × 10-34Js
p = 4.04 × 10-24 Kgms-1
So putting the values in equation we get
So the de Broglie wavelength of electron is 0.164 nm
Note:we could have also used the direct relation when an electron is accelerated through a potential difference of V volts its de Broglie wavelength is given by
1Ao = 10-10m
Question 13.
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
Answer:
Here we are given kinetic energy of electron
K = 120 eV
We know 1 eV = 1.6 × 10-19 J
i.e. kinetic Energy ,K = 120 × 1.6 × 10-19 J = 1.92 × 10-17J
(a) We know kinetic energy of a particle is given by the relation
Where K is the kinetic energy of a particle of mass m moving with speed v
Now multiplying L.H.S. and R.H.S. of the equation by m we get
or 
But we know momentum is given by relation, p = mv
So, substituting we get
or 
i.e. p = √2mK
Or we can say momentum P of any particle can be expressed in terms of its mass m and kinetic energy K as
p = √2mK
Particle is electron so we have mass of electron
m = 9.1 × 10-31Kg
kinetic energy of particle
K = 1.92 × 10-17J
Putting the values in equation we get
= 5.91 × 10-24 Kgms-1
So we get momentum of electron is 5.91 × 10-24 Kgms-1
(b) But we know momentum is given by relation
P = mv
Where P is momentum of particle of mass m moving with speed v
So we get speed of electron is
v = P/m
here the momentum of electron
P = 5.91 × 10-24 Kgms-1
Mass of electron
m = 9.1 × 10-31Kg
so putting these values we get the speed of electron as
or we can say speed of electron is 6.5 × 106 ms-1
(c) Now we know de Broglie wavelength of a Particle is given by relation
𝜆 = h/mv
Where 𝜆 is de Broglie wavelength of the particle having mass m and moving with velocity v and h is Planck’s constant
But we know momentum is given by relation
P = mv
Where P is the momentum of particle having mass m and moving with velocity v
So substituting we get de Broglie wavelength of Particle as
𝜆 = h/P
value of Planck’s constant
h = 6.6 × 10-34Js
and momentum of electron is
P = 5.91 × 10-24 Kgms-1
Putting these values we get
= 0.112 nm
So the wavelength of the electron is 0.112 nm
Question 14.
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron, and
(b) a neutron, would have the same de Broglie wavelength.
Answer:
Now we know de Broglie wavelength of a Particle is given by relation
𝜆 = h/mv
Where 𝜆 is de Broglie wavelength of the particle having mass m and moving with velocity v
But we know momentum is given by relation
p = mv
So, substituting we get de Broglie wavelength of Particle as
𝜆 = h/p
Or we can say momentum of particle is
p = h/ 𝜆
We know kinetic energy of a particle is given by the relation
Where K is the kinetic energy of a particle of mass m moving with velocity v
Now multiplying L.H.S. and R.H.S. of the equation by m we get
or
But we know momentum is given by relation
P = mv
So substituting we get
or 
we get kinetic energy of particle K in terms of momentum P and mass of particle m
so putting P = h/ 𝜆
we get kinetic energy of particle as
Where h is the Planck’s constant, m is mass of particle having de Broglie wavelength 𝜆
(a) Now we are given an electron whose de Broglie Wavelength should be
𝜆 = 589 nm = 589 × 10-9 m
We know mass of electron is, m = 9.1 × 10-31 Kg
and value of Planck’s constant, h = 6.6 × 10-34Js
putting values in the relation 
to find the kinetic energy of electron K we get
So kinetic energy of the electron is 6.9 × 10-25J
Converting it to eV
1 eV = 1.6 × 10-19J
i.e.
1J = (1/1.6 × 10-19)eV
So we get kinetic energy of electron as
Or we get Kinetic energy of electron as 4.31𝜇eV
(b) Now we are given an neutron whose de Broglie Wavelength should be
𝜆 = 589 nm = 589 × 10-9 m
We know mass of electron is
m = 1.66 × 10-27 Kg
and value of Planck’s constant
h = 6.6 × 10-34Js
putting values in the relation to find the kinetic energy of neutron K we get
So kinetic energy of the electron is 3.78 × 10-28J
Converting it to eV
We know
1 eV = 1.6 × 10-19J
i.e.
1J = (1/1.6 × 10-19)eV
So we get kinetic energy of electron as
Or we get Kinetic energy of electron as 2.36 neV
Question 15.
What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 × 10–9 kg drifting with a speed of 2.2 m/s?
Answer:
Now, we know de Broglie wavelength of Particle can be given by relation
Where, 𝜆 is de Broglie wavelength of a Particle
m, mass of the body
v, velocity of the body
h, planck constant = 6.63 × 10-34 Js
(a)
given,
mass of bullet, m = 0.040 kg
the speed of bullet is, v = 1.0 km/s = 1000 m/s
putting the valued of m, v, h in the relation
we get the de Broglie wavelength of Bullet as
So, de Broglie wavelength of Bullet is 1.65 × 10-35 m
(b)
given,
mass of Ball m = 0.060 kg
the speed of Ball is, v = 1.0 m/s
putting the valued of m , v, h in the relation
we get the de Broglie wavelength of Ball as
So, de Broglie wavelength of Ball is 1.1 × 10-32 m
(c)
given,
mass of dust particle m = 1 × 10-9 kg
the speed of dust particle is
v = 2.2 m/s
putting the valued of m, v, h
we get the de Broglie wavelength of Ball as
So, de Broglie wavelength of Dust Particle is 3.0 × 10-25 m
Question 16.
An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
Answer:
(a) In order to find momentum, we know de Broglie wavelength for any particle is given by
𝜆 = h/P
Or P = h/𝜆
Where h is Planck’s constant
h = 6.63 × 10-34 Js
and P is the momentum of particle
here we are given the de Broglie wavelength
𝜆 = 1.00 nm = 1.00 × 10-9 m
Note:we can see that momentum only depends on de Broglie wavelength and Planck’s Constant which is same for both electron or Photon(Light Particle) so the momentum for both will also be same
Putting values we get momentum
So momentum of electron and photon is 6.63 × 10-25Kgms-1
(b) we know energy of an photon is given by the relation
E = hc/𝜆
Where E is the energy of Photon having wavelength 𝜆 and speed c,h is Planck’s constant
h = 6.63 × 10-34 Js
Speed of light or photon in free space
c = 3 × 108 m/s
And we are given wavelength
𝜆 = 1.00 nm = 1.00 × 10-9 m
Putting these values we get
Converting it to eV
We know
1 eV = 1.6 × 10-19J
i.e.
1J = (1/1.6 × 10-19)eV
So we get energy of Photon as
so we get energy of Photon as 1.24KeV
(c) We know kinetic energy of a particle is given by the relation
Where K is the kinetic energy of a particle of mass m moving with velocity v
Now multiplying L.H.S. and R.H.S. of the equation by m we get
or
But we know momentum is given by relation
P = mv
Where m is mass of Particle moving with velocity v
So substituting we get
or we get kinetic energy of particle K in terms of momentum P and mass of particle m as
The particle is electron and mass of electron is
m = 9.1 × 10-31Kg
Here the momentum of electron
P = 6.63 × 10-25Kgms-1
Putting these values in above equation we have
Converting it to eV
We know
1 eV = 1.6 × 10-19J
i.e.
1J = (1/1.6 × 10-19)eV
So we get Kinetic energy of electron as
so we get Kinetic energy of electron as 1.51eV
Question 17.
For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10–10m?
Answer:
Now we know de Broglie wavelength of a Particle is given by relation
𝜆 = h/mv
Where 𝜆 is de Broglie wavelength of the particle having mass m and moving with velocity v
But we know momentum is given by relation
P = mv
Where P is the momentum of particle having mass m and moving with velocity v
So substituting we get de Broglie wavelength of Particle as
𝜆 = h/P
Or we can say momentum of particle is
P = h/ 𝜆
We know kinetic energy of a particle is given by the relation
Where K is the kinetic energy of a particle of mass m moving with velocity v
Now multiplying L.H.S. and R.H.S. of the equation by m we get
or
But we know momentum is given by relation
P = mv
So substituting we get
or
we get kinetic energy of particle K in terms of momentum P and mass of particle m
so putting P = h/ 𝜆
we get kinetic energy of particle as
Where h is the Planck’s constant, m is mass of particle having de Broglie wavelength 𝜆
Here,
The de Broglie Wavelength is, 𝜆 = 1.40 × 10–10m
The particle is neutron and mass of neutron is, m = 1.66 × 10–27Kg
putting these values we get the kinetic energy of neutron as
So kinetic energy of the neutron is 6.75 × 10-21J
Converting it to eV
We know
1 eV = 1.6 × 10-19J
i.e.
1J = (1/1.6 × 10-19)eV
So we get kinetic energy of neutron as
Or we get Kinetic energy of neutron as 0.042 Ev
Question 18.
Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K.
Answer:
Here the neutron is in thermal equilibrium with matter and it have a kinetic due which depends upon temperature of surrounding
We have a relation to find kinetic energy of a atomic particle at any Temperature which is
Where EK is the kinetic energy of particle at temperature T and K is Boltzmann constant
K = 1.38 × 10-23 JK-1
Here temperature is
T = 300 K
So linetic energy of neutron is
So kinetic energy of neutron is
Ek = 6.21 × 10-21J
And we know Kinetic energy of a particle is given by the relation
Where vis the speed of the particle and m is the mass of the particle
Multiplying both side sides by m
Or 
So we get
We know de-Broglie wavelength of a particle is given by relation
where 𝜆 is the de-Broglie wavelength of a particle having mass m and moving with velocity v, h is Planck’s constant
putting the value of
in above equation we get de Broglie wavelength
Here the particle is neutron and we know mass of neutron
m = 1.66 × 10-27Kg
Value of Planck’s constant
h = 6.63 × 10-34Js
and kinetic energy of neutron is
Ek = 6.21 × 10-21J
Putting values in above equation we get
Or 𝜆 = 0.146 × 10-9m = 0.146 nm
So the de Broglie wavelength of neutron is 0.146 nm
Question 19.
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Answer:
we know the relation between energy and momentum P of a electromagnetic radiation having velocity C and energy E is
P = E/C
and we know energy of electromagnetic radiation is given by relation
E = h𝜈
Where E is the energy of Photon, 𝜈 is the frequency of electromagnetic radiation and h is Planck’s constant
So putting value of E in equation P = E/c
we get
P = h𝜈/c
We know relation between frequency wavelength and velocity of electromagnetic radiation as
c = 𝜈𝜆
Where 𝜈 is the frequency and 𝜆 is wavelength of electromagnetic radiation having velocity c
So re arranging we get
𝜈 = c/𝜆
Putting the value of 𝜈 in equation P = h𝜈/c
We get
P = hc/λc = h/𝜆
Or we can say the wavelength of electromagnetic radiation is given as
𝜆 = h/P
Where h is Planck’s constant and P is momentum of electromagnetic radiation
Now we know de Broglie wavelength of quantum (photon) can be given by relation
𝜆 = h/mv
Where 𝜆 is de Broglie wavelength of a photon having mass m and moving with velocity v (here the velocity of photon v = C)
But we know momentum is given by relation
P = mv.
So, substituting we get de Broglie wavelength of quantum (photon) as
𝜆 = h/P
This is same as wavelength of electromagnetic radiation
Question 20.
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Answer:
We have a relation to find kinetic energy of a gas molecule at any Temperature which is
Where EK is the kinetic energy of a particle at temperature T and K is boltzmann constant
K = 1.38 × 10-23 JK-1
And we know Kinetic energy of a particle is given by the relation
Where vrms is the root meansquare speed of the particle and m is the mass of the particle
So equating both equations we have
On solving we get the root meansquare speed of the particle as
Where Vrms is the root mean square speed of the particle of mass m at temperature T and K is boltzmann constant
K = 1.38 × 10-23 JK-1
Here the temperature is
T = 300K
We are given the particle is a nitrogen molecule (N2) mass of the Nitrogen atom is
mN = 14.0076 u
so mass of N2 molecule is
m = 2 × 14.0076 u = 28.0152 u
converting it to Kg
we know 1 u = 1.66 × 10-27 Kg
so mass of Nitrogen molecule is
m = 28.0152 × 1.66 × 10-27Kg = 46.50 × 10-27Kg
= 4.650 × 10-26Kg
We know de-Broglie wavelength of a particle is given by relation
where 𝜆 is the de de-Broglie wavelength of a particle having mass m and moving with velocity v , here velocity is the root mean square speed of the particle, h is Planck’s constant
h = 6.63 × 10-34 Js
putting the value root mean square speed
in above equation we get
Solving we get the de Broglie wavelength is
Putting value of m,T,K and h in above equation we get
Or 𝜆 = 0.028 × 10-9m = 0.028 nm
So, the de Broglie wavelength of Nitrogen Molecule is 0.028 nm
Additional Exercises
Question 1.(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter.
Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 × 1011 C kg–1.
(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?
Answer:
Given:
Potential difference between collector and emitter = 500V
Specific charge of electron (charge per unit mass e/m) = 1.76 × 1011 C
Kinetic energy of an electron is given by:
…(1)
Where,
M = mass of electron
v = velocity of electron
e = charge of electron
V = potential difference (accelerating potential)
(a) From equation (1), we can write
…(2)
By putting the values in equation (2) we can find electron velocity.
v = 1.327 × 107 ms-1
(b) Accelerating potential, V = 10MV = 106V
Let speed of electron be v1
Again putting the values in equation (2),
v1 = 
v1 = 1.8 × 109ms-1
This result is wrong as we understand that speed of light
(i.e. 3 × 108 ms-1) is the theoretical limit of the speed.
Such problems can be dealt using relativistic mechanics,
Relativistic mass is given by:
m = 
Where,
m = relativistic mass
m0 = rest mass
v = velocity of particle
c = speed of light
At relativistic speeds, kinetic energy is given by,
KE = mc2-m0c2
Question 2.
(a) A monoenergetic electron beam with electron speed of 5.20 × 106 m s–1 is subject to a magnetic field of 1.30 × 10–4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011C kg–1.
(b) Is the formula you employ in (a) valid for calculating the radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
[Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasize the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]
Answer:
Given:
Speed of electron, v = 5.20 × 106 ms-1
Magnetic field strength normal to the beam = 1.30 × 10–4 T
Charge to mass ratio (e/m) = 1.76 × 1011 C Kg-1
(a) Force applied by magnetic field,
F = e|v × B|
F = e × v × B × sinθ …(1)
Where,
e = Charge on electron
v = velocity of particle
B = magnetic field strength
θ = angle between Magnetic field and velocity
Since the electron traces a circular path, we can use the following equation of centrifugal force:
F = 
…(2)
Where,
m = mass of particle
v = velocity of particle
r = radius of circle traced
By equating (1) and (2), we get,
evB × sin90° =
r = 
→ r = 
…(3)
→ r = 
→ r = 22.7 6 cm
(b) Energy of the electron beam, E = 20 MeV
E = 20 × 106 × 1.6 × 10-19J
The energy of an electron is given by:
E = 
From the above equation, we can write,
v = 
…(4)
putting the value in equation (4)
v = 
v = 2.6 × 109 ms-1
This result is incorrect because the speed of any massive object can’t exceed the speed of light (i.e. 3 × 108 ms-1). We can’t use equation (4) in the case where the speed is relativistic:
At relativistic speeds, mass is given by,
m =
Where,
m = relativistic mass
m0 = rest mass
v = velocity of particle
c = speed of light
the radius of the path traced is given by,
r = …(3)
By substituting the value of relativistic mass in equation (3) we get,
r = 
…(4)
By using equation (4) we can find the radius traced by electrons moving at relativistic speed.
By substituting value of relativistic mass in equation
Question 3.
An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (∼10–2 mm of Hg). A magnetic field of 2.83 × 10–4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data.
Answer:
Given:
Potential at collector, V = 100V
Magnetic field strength, B = 2.83 × 10-4 T
Radius of orbit traced by electron, r = 12.0 cm
Let specific charge (i.e. e/m) = K
Kinetic energy of an electron is given by:
KE = = eV
→ v = 
…(1)
Where,
M = mass of electron
v = velocity of electron
e = charge of electron
V = potential difference (accelerating potential)
We also know that the bending is caused by the magnetic field,
So we can write,
Magnetic force = centrifugal force
e × v × B =
→ e × B = 
→ v = 
…(2)
Where,
e = Charge on electron
v = velocity of particle
B = magnetic field strength
m = mass of electron
r = radius of trajectory
By equating (1) and (2) we can write,
.
→
By putting the values in above equation we get,
Hence the specific charge is 1.7 × 10-11 C Kg-1.
Question 4.
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
Answer:
Given:
The wavelength of radiation produced by tube, λ = 0.45 Å
i.e. 0.45 × 10-10 m
(a) The maximum energy of the photon is given by,
…(1)
Where,
E = Energy of photon
h = Planck’s constant = 6.6 × 10-34 Js
c = 3 × 108 ms-1
λ = wavelength of photon
by putting the values in equation (1), we get,
E = 44 × 10-16 J or 27.6 × 103 eV
The maximum energy of the photon in x-ray is 44 × 10-16J.
(b) For a photon to have 27 KeV of energy, the accelerating potential must be of the order of 30 KeV.
Question 5.
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as the annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1BeV = 109 eV)
Answer:
Given:
Total energy of system, E = 10.2 BeV = 10.2 × 109 × 1.6 × 10-19J
E = 16.32 × 10-10J
Energy in each ray, E’ = E/2
E’ = 8.16 × 10-10J
We know that Energy in a photon is given by,
…(1)
Where,
E = Energy of photon
h = Planck’s constant = 6.6 × 10-34 Js
c = 3 × 108 ms-1
λ = wavelength of photon
From equation (1) we can get,
λ = 2.34 × 10-16m
Hence, the wavelength of each ray is 2.3 × 10-16 m.
Question 6.
Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in the barely detectable light.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radio waves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (∼10–10 W m–2). Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about 6 × 1014 Hz.
Answer:
Given:
Power of transmitter, E’ = 10 KW = 10000 Js-1
Wavelength of radio waves emitted = 500 m
(a) We know that Energy in a wave is given by,
…(1)
Where,
E = Energy of photon
h = Planck’s constant = 6.6 × 10-34 Js
c = 3 × 108 ms-1
λ = wavelength
E = 3.9 × 10-28 J
Total power = number of photons emitted × Energy of photon
E’ = n × E
n = 
n = 2.55 × 1031 s-1
n = 3 × 10-31 s-1
Here we see that the number of radio waves emitted per second is very high.
(b) Given:
Intensity of light perceived by human eyes, I = 10-10Wm-2
Area of pupil, A = 0.4 × 10-4 m2
Frequency of white light, v = 6 × 1014Hz
The energy of each photon is given by,
E = hv
Where,
E = energy of photon
h = Planck’s constant = 6.6 × 10-34Js
v = 6 × 1014Hz
E = 6.6 × 10-34Js × 6 × 1014 s-1
E = 3.96 × 10-19J
Energy of each photon is 3.96 × 10-19J
Let total number of photons being emitted per second, falling on unit area = n
We define intensity as the amount of energy falling on unit area in unit time so we can write,
I = n × 3.96 × 10-19J
10-10 Jm-2s-1 = n × 3.96 × 10-19J
n = 2.52 × 108 m-2s-1
Number of photons entering pupil = area of pupil × n
Number of photons entering pupils = 0.4 × 10-4m2 × 2.52 × 108 m-2s-1
→ np = 1.0082 × 104s-1
Almost 10000 photons enter our pupil per second.
Question 7.
Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is –1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (∼105 W m–2) red light of wavelength 6328 Å produced by a He-Ne laser?
Answer:
Given:
Wavelength of light, λ = 2271 Å = 2271 × 10-10m
Power of mercury source, E = 100 Js-1
Stopping potential, Vs = -1.3 V
Let frequency of light = v
Work function, Φ0 is given by,
Φ0 = hv-eVs
Φ0 = 
Where,
h = Planck’s constant = 6.6 × 10-34Js
c = speed of light = 3 × 108m
λ = wavelength of light
e = charge on each electron = 1.6 × 10-19C
Φ0 = 
Φ0 = 6.64 × 10-19J
Φ0 = (6.64/1.6) × 10-19 eV = 4.15 eV
Let v0 be the threshold frequency of the metal,
Φ0 = hv0
v0 = Φ0/h
→ v0 = 6.6 × 10-19/ 6.6 × 10-34
→ v0 = 1.00. × 10-15 s-1
Wavelength of red light, λ’ = 6323 × 10-10m
Frequency of red light can be given as,
v' = c/λ’
v’ = 
v’ = 4.74 × 1014 Hz
Since the threshold frequency is greater than the frequency of red light, the photocell will not respond to the red light produced.
Question 8.
Monochromatic radiation of wavelength 640.2 nm (1nm = 10–9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Answer:
Given:
Wavelength of light, λ = 640.2 nm = 640.2 × 10-9m
Stopping potential, V0 = 0.54 V
Let frequency of light = v
Work function, Φ0 is given by,
Φ0 = hv-eVs
Φ0 = …(1)
Where,
h = Planck’s constant = 6.6 × 10-34Js
c = speed of light = 3 × 108m
λ = wavelength of light
e = charge on each electron = 1.6 × 10-19C
Φ0 = 
Φ0 = 2.229 × 10-19J or 1.39 eV
Wavelength of radiation emitted by iron, λ’ = 427.2 nm
λ’ = 4.27.2 × 10-9m
The new stopping potential can be found using equation (1),
Let new stopping potential = v’
By putting the values in equation (1) we get,
Φ0 = 
→ 
→ 
→ v’ = 1.5 eV
Therefore, the new stopping potential is 1.5 eV.
Question 9.
A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:
λ1 = 3650 Å, λ2 = 4047 Å, λ3 = 4358 Å, λ4 = 5461 Å, λ5 = 6907 Å,
The stopping voltages, respectively, were measured to be:
V01 = 1.28 V, V02 = 0.95 V, V03 = 0.74 V, V04 = 0.16 V, V05 = 0 V
Determine the value of Planck’s constant h, the threshold frequency and work function for the material.
[Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 × 10–19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]
Answer:
Given:
The frequency, v of a wave with wavelength λ is given by,
v = c/λ …(1)
(c = speed of light = 3 × 108ms-1)
By using equation (1) we can find frequency for each case,
The equation for photo-electric effect is given as,
Φ0 = hv- eV0
We can rewrite this equation as,
V0 = 
…(2)
Where,
h = Planck’s constant
c = speed of light = 3 × 108m
λ = wavelength of light
e = charge on each electron = 1.6 × 10-19C
We find that the slope of the graph remains same,
Slope of the line = 
From equation (2),
We can write,
Slope = h/e
h = e × slope
h = 
h = 6.58 × 10-34 Js
From the same graph, we can infer that, threshold frequency of the metal is 5 × 1014 Hz
i.e. v0 = 5 × 1014Hz
Work function, Φ0 = hv0 …(2)
Where,
h = Plank’s constant = 6.57 × 10-34Js
v0 = threshold frequency
By plugging the data in equation (2), we get,
Φ0 = 6.57 × 10-34Js × 5 × 1014Hz
→ Φ0 = 3.28 × 10-19 J
→ 
→ Φ0 = 2eV
Hence, the work function of the metal is 2eV.
Question 10.
The work function for the following metals is given:
Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?
Answer:
Given:
Wavelength of radiation from laser = 3300 Å = 3300 × 10-10m
Energy radiated by the laser, E is given by,
…(1)
Where,
E = Energy of photon
h = Planck’s constant = 6.6 × 10-34 Js
c = 3 × 108 ms-1
λ = wavelength of photon
putting values in equation (1) we get,
E = 6 × 10-19 J
E = (6/1.6) × 10-19
E = 3.158 eV
We observe that Na and K have work functions less than the energy radiated by the laser so photoelectric effect will take place in case of them, whereas no photo-electric effect will occur in case of Mo and Ni.
Even if the laser is brought closer, it will have no effect on the outcome as the photo-electric effect is independent of the intensity of light.
Question 11.
Light of intensity 10–5 W m–2 falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Answer:
Given:
Intensity of light, I = 10-5 Wm-2
Surface area of photo-cell, A = 2 cm2 = 0.0002 m2
Work function of sodium, Φ0 = 2eV = 2 × 1.6 × 10-19 = 3.2 × 10-19J
By knowing the effective area of each sodium atom we can find the absorption of incident energy,
Effective area of each sodium atom, ANa = 10-20 m2
Number of layers of sodium, n = 5
Number of atoms absorbing radiation, NNa = n × (ANa/ A)
→ NNa = 5 × (10-20 m2/ 0.0002 m2)
→ NNa = 1017
So we conclude that 1017 atoms are effectively absorbing radiation.
Energy absorbed per atom, E = I/NNa
E = 10-5Js-1/1017
→ E = 2 × 10-26 Js-1
Times required for photo electric emission, t is give by,
t = Φ0/E
t = 3.2 × 10-19 J / 2 × 10-26 Js-1
t = 1.6 × 107 s
t = 0.5 years
Hence the time requires to initiate photoelectric emission is 0.5 years which is impractical hence, the wave model stands in disagreement with the experimental results.
Question 12.
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (me = 9.11 × 10–31 kg).
Answer:
Given:
Wavelength of probe, λ = 1 Å = 10-10m
Mass of electron, me = 9.11 × 10-31Kg
The kinetic energy of an electron is given by:
We can write,
mv = (2 × m × KE)0.5
P = (2 × m × KE)0.5 ...(1)
Where,
m = mass of electron
v = velocity of electron
p = momentum of particle
De-broglie wavelength is given by,
Where,
λ = Wavelength
h = Planck’s constant
p = momentum
From equation (1) we can write,
…(2)
From equation (2) we can write that,
KE = 
Putting the values in above equation we get,
KE = 
KE = 2.39 × 10-17 J
KE for an electron is 2.39 × 10-17 J.
Now for photon,
E’ = 
…(4)
Where,
E = energy of photon
h = Planck’s constant
c = speed of light
λ = wavelength of the photon
By putting values in equation (4 ) we get,
E’ = 19.6 × 10-16 J
The photon will have more energy as compared to the accelerated electron for the same wavelength.
Question 13.
Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain.
(mn = 1.675 × 10–27 kg)
Answer:
Given:
Kinetic energy of neutron, E = 150 eV = 150 × 1.6 × 10-19
E = 2.4 × 10-17 J
Mass of neutron, mn = 1.675 × 10-27 Kg
Kinetic energy of a particle is given by,
We can write,
mv = (2 × m × KE)0.5
P = (2 × m × KE)0.5 ...(1)
Where,
m = mass of electron
v = velocity of electron
p = momentum of particle
By putting the equation (1) in de- Broglie equation, we get,
…(2)
Now substituting the values in eq (2) we get,
λ = 2.327 × 10-12m
A neutron cannot be used in diffraction experiment as lattice spaces are of the order of few 10-10 m, whereas the wavelength of a neutron beam is 100 times smaller.
Question 14.
Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27°C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
Answer:
De- Broglie wavelength, λ = 1.447 × 10-10 m
Room temperature, T = 300K
The average KE of neutron is given as,
KE = 3/2 KT …(3)
Where,
K = Boltzmann constant = 1.38 × 10-23 Jmol-1K-1
From equation 2,
Using equation (3) we can write,
By putting values in above equation we get,
λ = 1.447 × 10-10 m
Now, the wavelength has come into comparable range with the inter-atomic spaces, it is now suitable for diffraction. Hence a beam should be thermalized before using for diffraction.
Question 15.
An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?
Answer:
Given:
Accelerating voltage, V = 50 KV
Energy of each electron is given by,
E = eV …(1)
Where,
e = Charge on electron = 1.6 × 10-19C
V = accelerating Voltage
Putting the values in equation (1)
E = 1.6 × 10-19C × 50 × 1000V
E = 8 × 10-15 J
De Broglie wave length is given by,
Where,
m = mass of electron
h = Planck’s constant = 6.6 × 10-19 Js
E = Energy
By putting the values in above equation we get,
λ = 5.4 × 10-12 m.
As compared to an optical telescope which uses yellow light, electron microscope has better resolving power (about 105 times better). As the resolving power is inversely proportional to the wavelength of light used.
Question 16.
The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10–15 m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)
Answer:
Given:
Order of length of quark structure, λ = 10-15m
Rest mass energy of electron = 0.511 MeV
i.e. m0c2 = 0.511 MeV
→ m0c2 = 0.511 × 1.6 × 10-19
→ m0c2 = 0.817 × 10-13 J
By using De-Broglie wavelength equation we can write,
Where,
p = momentum of particle
h = Planck’s constant = 6.6 × 10-34Js
λ = wavelength
The equation of energy at relativistic speed is,
E2 = p2c2 + m20c4
E2 = (6.6 × 10-19 × 3 × 108)2 + (0.817 × 10-13)2
→ E = (392.07 × 10-22)0.5
→ E = 1.9 × 10-10J
Hence the energy of electron emitted from linear accelerator is 1.9 × 10-10J.
Question 17.
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.
Answer:
Given:
Pressure, P = 1 atm = 101325 Pa
Temperature, T = 27° = 300K
De-Broglie wavelength is given by,
…(1)
Where,
h = Planck’s constant = 6.6 × 10-34 Js
m = mass of He
E = total energy
Mass of helium, m = Atomic mass/ number of atoms
Let there be one mole of He,
mass of 1 mol of He = 4 g
Then,
m = 4/ 6.023 × 1023
m = 6.64 × 10-24 = 6.64 × 10-27Kg …(2)
We know that average energy at temperature T is given by,
E = 3/2 kT …(3)
Where,
k = Boltzmann constant = 1.38 × 10-23 Jmol-1K-1
T = absolute temperature
Using the equation (3), we can rewrite equation (1) as,
…(4)
Putting the values in equation (4) we get,
λ = 0.72 × 10-10m
Using ideal gas equation we have,
PV = RT
PV = kNT
Where,
V = volume
N = number of moles
P = pressure
k = Boltzmann constant
T = absolute temperature
Mean distance between, r is given by,
r = 3.3 × 10-9m
The mean separation between the atoms is greater than the De- Broglie wavelength.
Question 18.
Compute the typical de Broglie wavelength of an electron in a metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10–10 m.
[Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This indistinguishability has many fundamental implications which you will explore in more advanced Physics courses.]
Answer:
Given:
Temperature, T = 27° C = 300 K
Mean separation between electrons, r = 2 × 10-10 m
De-Broglie wavelength of electron is given by,
…(1)
Where,
h = Planck’s constant = 6.6 × 10-34 Js
m = mass of electron = 9.1 × 10-31 Kg
k = Boltzmann constant = 1.38 × 10-23 Jmol-1K-1
T = absolute temperature
Putting the values in equation(1), we get,
λ = 6.2 × 10-9 m
The De-Broglie wavelength is much larger than the interelectronic separation.
Question 19.
Answer the following question:
Quarks inside protons and neutrons are thought to carry fractional charges [( + 2/3)e ; (–1/3)e]. Why do they not show up in Millikan’s oil-drop experiment?
Answer:
Quarks inside protons and neutrons are known to have fractional charges. Quarks do not show up in Millikan's oil drop experiment as these fractional charges do not exist independently, rather held together by strong nuclear force. These fractional charges are not found free in nature.
Question 20.
Answer the following question:
What is so special about the combination e/m? Why do we not simply talk of e and m separately?
Answer:
e/m or the specific charge is a important ratio as it is found in basic equations of electro-magnetism, for instance,
evB = mv2/r …(1)
→ (e/m)vB = v2/r
v = Br(e/m)
eV = 1/2mv2 …(2)
→ v = 
We see that the ratio (e/m) has an impact on the dynamics of electron in magnetic field hence e/m is given more importance that e or m.
Question 21.
Answer the following question:
Why should gases be insulators at ordinary pressures and start conducting at very low pressures?
Answer:
At atmospheric/ ordinary pressures the Ions of gases move randomly and collide with each other , this random motion prevents it from setting a definite current in given direction, at low pressures these random motions decrease and if a potential difference is created between two points the gaseous Ions are able to move in definite direction setting a current. Hence, the gases become conductor at very low pressures.
Question 22.
Answer the following question:
Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
Answer:
Work function is the minimum energy required to set off the photo-electric emission. As the incident radiation falls on electrons they absorb defined wavelengths and if they gain enough energy they jump out of metal. But the energy absorption depends is very specific and depends on the initial energy level of electrons. Since all electrons are not in same energy level, all of them can’t absorb the monochromatic wavelength. Since different electrons start from different energy levels so there a energy distribution of photo-electrons.
Question 23.
Answer the following question:
The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:
But while the value of λ is physically significant, the value of ν (and therefore, the value of the phase speed ν λ) has no physical significance. Why?
Answer:
We understand that momentum and wavelength of photons have physical significance as it is evident from X-diffraction crystallography. But Energy of photon which is related to the frequency is arbitrary to within an additive constant, hence absolute value of frequency has no physical significance.