Atoms
Class 12th Physics Part II CBSE Solution
Exercises
Question 1.Choose the correct alternative from the clues given at the end of the each statement:
(a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no different from/much less than.)
(b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson’s model/ Rutherford’s model.)
(c) A classical atom based on .......... is doomed to collapse. (Thomson’s model/ Rutherford’s model.)
(d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (Thomson’s model/ Rutherford’s model.)
(e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford’s model/both the models.)
Answer:
(a) The correct alternative is: no different from
Explanation: The size of the atom in the thomson’s model and the size of the atom in Rutherford’s model of an atom were of same magnitude.
(b) The correct alternative is: Thomson’s model, Rutherford’s model
Explanation: This is because when the electrons are in the ground state in case of Thomson’s model, they are in stable equilibrium whereas in case of Rutherford’s model of an atom, when electrons are at ground state, they experience a net force.
(c) The correct alternative is: Rutherford’s model
Explanation: A classical atom based on Rutherford’s model is doomed to collapse.
(d) The correct alternative is: Thomson’s model, Rutherford’s model
Explanation: The mass distribution of an atom is nearly continous in case of Thomson’s model of an atom, whereas in case of Rutherford’s model of an atom, the mass distribution is not uniform or continuous.
(e) The correct alternative is: both the models.
Explanation: In both Rutherford and Thomson’s model of an atom, the most of the mass is due to the positive part of the atom i.e. protons.
Question 2.
Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.)
What results do you expect?
Answer:
In alpha-particle scattering experiment, if we use a thin sheet of solid hydrogen in place of the gold foil, then the mass of Hydrogen will be lesser than that of incident alpha particle and the scattering angle will be not large enough. As mass of the scattering particle is more than the target nucleus, the alpha particles will not bounce while using solid Hydrogen.
Question 3.
What is the shortest wavelength present in the Paschen series of spectral lines?
Answer:
The shortest wavelength present in the Paschen series of spectral lines is given for n1 = 3 and n2 = ∞
The diagram is shown below:
Rydberg’s formula is given as:
where
Z is atomic number
R is Rydberg’s constant, R = 1.9074 x 107 m-1
h is planck’s constant and is equal to 6.64 × 10-34 Js
c is speed of light and is equal to 3.0 × 108 m/s
substituting the values we get,
⇒ 
⇒ 
⇒ 
⇒ 
⇒ λ = 8.2012 × 10-7 m ~ 820.12 × 10-9 m ~
Or λ = 820 nm
(∵ 10-9 m = 1nm)
Question 4.
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?
Answer:
Given: energy = 2.3 eV
In joules, Energy E = 2.3 × 1.6 × 10-19 = 3.68 × 10-19J
Energy is given as:
E = hv
Where h is Planck’s constant
v is frequency of the radiation
Or 
On calculating, we get
Frequency v =0.55 × 10-15
Or Frequency v = 5.55 × 10-14 Hz.
Question 5.
The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?
Answer:
Given: Total energy of the Hydrogen atom = Ground state energy of hydrogen atom is –13.6 eV
Since, the kinetic energy is equal to the negative of the total energy
∴ Kinetic energy of the hydrogen atom = - (-13.6 eV) = 13.6 eV
The potential energy is equal to the twice of the negative of the total energy.
∴ Potential energy of hydrogen atom =-2 × - (-13.6 eV)
Potential energy = -27.2 eV.
Question 6.
A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
Answer:
For ground level: n1 =4
Energy for the ground level of hydrogen atom, E1 = -13.6/n12
Substituting the value, we get
E1 = -13.6/12 = -13.6 eV
When the hydrogen atom is excited to n2= 4, then the energy absorbed by the photon is given as:
E2 = -13.6/42
E2= -13.6/16
The energy absorbed by the photon can be calculated by the following:
E = E2- E1
Or, E = (-13.6/16)- (-13.6/1)
On calculating, we get
E = 12.75 eV
In Joules, Energy is = 12.75 × 1.6 × 10-19 Joules
Or E = 2.04 × 10-18 J
Energy is given by the relation, E =hc/λ
Or λ = hc/E
Where h is Planck’s constant and is equal to 6.64 × 10-34 Js
c is speed of light and is equal to 3.0 × 108 m/s
substituting the values we get
On calculating, we get
λ = 9.7 × 10-8 m
or λ = 97 nm
Frequency can be calculated as follows:
v = c/λ
substituting the values, we get
v= 3.1 × 1015 Hz
Question 7.
Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels.
Answer:
The orbital speed of an electron in a hydrogen atom, is given by the relation
v1 = 
or v1 = 
where ε0 is the absolute permittivity of free space and is given as = 8.854 × 10-12 N-1C2m-2
h is Planck’s constant and is given as 6.64 × 10-34 Js
substituting the values, we get
v1 = 
On calculating, we get
v1 = 0.0218 × 108 m/s
Or v1 = 2.18 × 106 m/s
For n = 2
or v2 = 
substituting the values, we get
On calculating, we get
v2 = 1.09 × 106 m/s
Similarly, for n = 3, we get
or v3 =
substituting the values, we get
On calculating, we get
v3 = 7.27 × 106 m/s
Question 8.
Calculate the orbital period in each of these levels.
Answer:
Orbital period can be calculated as:
T1 = (2πr1)/v1
Where r1is the radius of the orbit and is given as follows:
Where ε0 is the absolute permittivity of free space and is given as = 8.854 × 10-12 N-1C2m-2
h is Planck’s constant and is given as 6.64 × 10-34 Js
e is charge on an electron and is equals to 1.6 × 10-19C
m is the mass of an electron, and is equals to 9.1 × 10-31 Kg
substituting the values, we get
⇒ 
On calculating, we get
T1 = 15.27 × 10-16 seconds
For n = 2, we get
⇒ 
On calculating, we get
T2 = 1.22 × 10-15 seconds
For n = 3, we get
⇒ 
On calculating, we get
T3 = 4.12 × 10-15 seconds.
Question 9.
The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10–11 m. What are the radii of the n = 2 and n =3 orbits?
Answer:
Given, radius of innermost orbit, r1 = 5.3 × 10–11 m
Then, let r2 be the radius of orbit at n = 2.
Then, r2 = (n)2r1
As shown in the diagram:
⇒ r2 = 22 × 5.3 × 10–11 m
⇒ r2 = 2.12 × 10–10 m
For n = 3, we have,
⇒ r3 = (n)2 r1
⇒ r3 = 32 × 5.3 × 10–11 m
⇒ r3 = 4.77 × 10–10 m
Hence, the radii of an electron for n = 2 and n=3 orbits are 2.12 × 10–10 m and 4.77 × 10–10 m
Question 10.
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Answer:
Energy of the electron used for bombardment of gaseous hydrogen at room temperature = 12.5 eV.
Energy of gaseous hydrogen at ground state = -13.6 eV
When gaseous hydrogen is bombarded with electron beam, then energy of hydrogen will be =
(-13.6 + 12.5) eV= -1.1 eV
Orbital energy is given by the relation E= -13.6/n2
For n = 3, orbital energy will be E= -13.6/32
E = - 1.5eV
Since the energy is equal to the energy of gaseous hydrogen. Therefore, the electron has made it transition from n= 1 to n= 3
During deexcitation of the hydrogen gas, the electron will return back from n =3 to n= 1 and forms a line of Lyman series of the hydrogen spectrum.
The wave number of the lyman series is given by the relation:
1/λ = Ry (1/12 – 1/n2)
Where Ry is the Rydberg’s constant and is equals to 1.097 × 107m-1
For n =3, substituting the values, we get
1/λ = 1.097 × 107 (m-1)[(1 – 1/32)]
On calculating, we get
λ = 102.55 nm
When electron makes transition from n = 2 to n =1 level
1/λ = 1.097 × 107 (m-1)[(1 – 1/22)]
On calculating, we get
λ = 121.54 nm
When the electron makes transition from n=3 to n = 2
1/λ = 1.097 × 107 (m-1)[(1/4 – 1/32)]
On calculating, we get
λ = 656.33 nm
∴ In Lyman series, two wavelengths 102.55 nm and 121.54 nm were emitted, and in Balmer series, one wavelength 656.33 nm is emitted.
Question 11.
In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3 × 104 m/s. (Mass of earth = 6.0 × 1024 kg.)
Answer:
Angular momentum is given by the relation:
⇒ 
Where M is the mass of Earth and is equal to 6.0 × 1024 kg.
v is the orbital speed of Earth = 3 × 104 m/s.
r is the orbital radius of Earth= 1.5 × 1011 m
n is the quantum number
substituting the values, we get
on calculating, we get
n =2.6 × 1074
Additional Exercises
Question 1.Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.
(a) Is the average angle of deflection of α-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
(b) Is the probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
(c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of α-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?
(d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of α-particles by a thin foil?
Answer:
(a) The average angle of deflection of α-particles by a thin gold foil predicted by Thomson’s model about the same as that predicted by Rutherford’s model. This is due to the reason that, in both Thomson and Rutherford model of an atom, the average angle of deflection was taken.
(b) The probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by Thomson’s model much less than that predicted by Rutherford’s model.
(c) With increase in the thickness of, the number of target atoms that will increase. This will lead to increase in the collisions, and as we know, scattering is due to single collision. Therefore, with increase in thickness, the chances of collisions increase.
Question 2.
The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10–40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.
Answer:
The radius of first Bohr orbit is given by the relation:
……………(1)
Where ε0 is the absolute permittivity of free space and is given as = 8.854 × 10-12 N-1C2m-2
h is Planck’s constant and is given as 6.64 × 10-34 Js
e is charge on an electron and is equals to 1.6 × 10-19C
me is the mass of an electron, and is equals to 9.1 × 10-31 Kg
mp is the mass of proton and is equals to 1.67 × 10-27 Kg
The forces of coulomb attraction between an electron and proton is given by the relation: 
Gravitational forces of attraction between an electron and proton is given by the relation
Where G is the Gravitational constant and is equal to G = 6.67 × 10-11 N m2 Kg2
When coulomb force and Gravitational forces of attraction are equal
Or 
…………(1)
Substituting the values of equation (1) in equation (2), we get
Substituting the values, we get
On calculating, we get
∴ r1= 1.21 × 1029 m
Since Universe is 150billion light years wide, which concludes that the radius of first Bohr orbit is much greater than the size of the universe.
Question 3.
Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n–1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.
Answer:
The energy of radiation at level n is given by the relation:
…(i)
Where v1 is the frequency of radiation at level n
H is Planck’s constant
m is mass of hydrogen atom
e is charge on electron
ε0 is permittivity of free space
The energy of radiation at level (n-1) is given by the relation:
……………(ii)
Where v2 is the frequency of radiation at level n-1
Due to deexcitation of the electrons, the energy released is:
E =E2- E1
hv = E2- E1…………………..(iii)
Putting equation (i) and (ii) in equation (iii), we get
v =
v = 
For large values of n, (n-1)≈n
v = 
Frequency of revolution of an electron is given by
Vc = v/2πr……………..(vi)
Velocity of electron in nth orbit is given by the following relation
…………………….(iv)
The radius of nth orbit is given by the following relation
r =
………………….(v)
Substituting the values of equation (iv) and (v) in equation (vi), we get
Question 4.
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10–10m).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value.
(b) You will find that the length obtained in A is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.
Answer:
(a) Given: charge on an electron and is equals to 1.6 × 10-19C
me is the mass of an electron, and is equals to 9.1 × 10-31 Kg
c is speed of light and is equal to 3.0 × 108 m/s
The quantity involving the above quantities is as follows:
1/4πε0 = 9 × 109Nm2C-2
Substituting the values we get,
On calculating, we get
= 2.81 × 10-15m
Therefore, the numerical quantity is much smaller than the size of the atom.
(b) Given: charge on an electron and is equals to 1.6 × 10-19C
me is the mass of an electron, and is equals to 9.1 × 10-31 Kg
Planck’s constant = 6.62 × 10-34Js
The quantity involving the above quantities is as follows:
1/4πε0 = 9 × 109Nm2C-2
Substituting the values we get
On calculating, we get
= 2.81 × 10-15m
The numerical quantity is of order of the atomic size.
Question 5.
The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?
Answer:
(a) Given: The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.
Kinetic energy of the electron in this state = negative of the total energy = -E
Kinetic energy of the electron in this state =-(-3.4)eV = + 3.4 eV
(b) Potential energy is given as the negative of the twice of the kinetic energy
U = -2 (3.4) eV
U = -6.8eV
(c) If the choice of the zero of potential energy is changed, then the value of potential energy of the system also changes and as we know the total energy is sum of kinetic energy as well as potential energy. Therefore, the potential energy will also changes.
Question 6.
If Bohr’s quantisation postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun?
Answer:
The angular momentum associated with planetary motion is large relative to the Planck’s constant. As the angular momentum of Earth in its orbit is of order of 1070 h which results in higher value of quantum level of order of 1070. The more is the value of n, the less will be the angular momentum and energies. Therefore, the quantum levels of planetary motion are considered as continuous.
Question 7.
Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (μ–) of mass about 207me orbits around a proton].
Answer:
Given: Mass of negatively charged muon mμ– = 207me
The Bohr’s radius is given by the relation:
re ∝ (1/mc)
Energy of ground state electronic hydrogen atom Ee∝ m
The value of the first Bohr orbit is given as re = 0.53A
Converting into metres, re =0.53 × 10-10m
Let us consider rμ is the radius of muonic hydrogen atom
At equilibrium
mμ rμ = me re
207me × rμ = me re
rμ = (0.53 × 10-10m)/207
rμ =2.56 × 10-13m
The ratio of the energies is given as:
Ee/Eμ = me/mμ = me/207me
We get
Eμ = 207EE
Substituting the value of energy, we get
Eμ = 207 × (-13.6)
Eμ =-2.81 keV