Bohr's Model and Physics of the Atom

Question 1:

Question 2:

Balmer series contains wavelengths ranging from 364 nm (for n₂ = 3) to 655 nm (n₂= $\infty$).
So, the given range of wavelength (380−780 nm) lies in the Balmer series.
The wavelength in the Balmer series can be found by
$\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$
Here, R =  Rydberg's constant = 1.097×10⁷ m^$-$1

The wavelength for the transition from n = 3 to n = 2 is given by
$$\begin{gathered} \frac{1}{{\lambda }_1}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ {\lambda }_1=656.3 \mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 4 to n = 2 is given by
$$\begin{gathered} \frac{1}{{\lambda }_2}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \\ {\lambda }_2=486.1 \mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 5 to n = 2 is given by
$$\begin{gathered} \frac{1}{{\lambda }_3}=R\left(\frac{1}{2^2}-\frac{1}{5^2}\right) \\ {\lambda }_3=434.0 \mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 6 to n = 2 is given by
$$\begin{gathered} \frac{1}{{\lambda }_4}=R\left(\frac{1}{2^2}-\frac{1}{6^2}\right) \\ {\lambda }_4=410.2 \mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 7 to n = 2 is given by
$$\begin{gathered} \frac{1}{{\lambda }_5}=R\left(\frac{1}{2^2}-\frac{1}{7^2}\right) \\ {\lambda }_5=397.0 \mathrm{nm} \end{gathered}$$

Thus, the wavelengths emitted by the atomic hydrogen in visible range (380−780 nm) are 5.
Lyman series contains wavelengths ranging from 91 nm (for n₂ = 2) to 121 nm (n₂ =$\infty$).
So, the wavelengths in the given range (50−100 nm) must lie in the Lyman series.
The wavelength in the Lyman series can be found by
$\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)$

The wavelength for the transition from n = 2 to n = 1 is given by
$$\begin{gathered} \frac{1}{{\lambda }_1}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\ {\lambda }_1=122 \mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 3 to n = 1 is given by
$$\begin{gathered} \frac{1}{{\lambda }_2}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\ {\lambda }_2=103 \mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 4 to n = 1 is given by
$$\begin{gathered} \frac{1}{{\lambda }_3}=R\left(\frac{1}{1^2}-\frac{1}{4^2}\right) \\ {\lambda }_3=97.3 \mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 5 to n = 1 is given by
$$\begin{gathered} \frac{1}{{\lambda }_4}=R\left(\frac{1}{1^2}-\frac{1}{5^2}\right) \\ {\lambda }_4=95.0 \mathrm{nm} \end{gathered}$$
The wavelength for the transition from n = 6 to n = 1 is given by
$$\begin{gathered} \frac{1}{{\lambda }_5}=R\left(\frac{1}{1^2}-\frac{1}{6^2}\right) \\ {\lambda }_5=93.8 \mathrm{nm} \end{gathered}$$

So, it can be noted that the number of wavelengths lying between 50 nm to 100 nm are 3.

Question 3:

The energy of hydrogen ion is given by
$E_{\mathrm{n}}=-\frac{(13.6 \mathrm{eV}) Z^2}{n^2}$
For the first excited state (n = 2), the energy of He⁺ ion (with Z = 2) will be $-$13.6 eV. This is same as the ground state energy of a hydrogen atom.
Similarly, for all the hydrogen like ions, the energy of the (n $-$ 1)th excited state will be same as the ground state energy of a hydrogen atom if Z = n.

Question 4:

As the electron collides, it transfers all its energy to the hydrogen atom.
The excitation energy to raise the electron from the ground state to the nth state is given by
$E=(13.6 \mathrm{eV})\times \left(\frac{1}{1^2}-\frac{1}{n^2}\right)$
Substituting n = 2, we get
E  = 10.2 eV
Substituting n = 3, we get
E'  = 12.08 eV

Thus, the atom will be raised to the second excited energy level.
So, when it comes to the ground state, there is transitions from n = 3 to n = 1.
Therefore, the wavelengths emitted will lie in the Lyman series (infrared region).

Question 5:

White radiations are x-rays that have energy ranging between 5-10 eV.
When white radiation is passed through a sample of hydrogen gas at room temperature, absorption lines are observed only in the Lyman series.
At room temperature, almost all the atoms are in ground state.
The minimum energy required for absorption is 10.2 eV (for a transition from n = 1 to n = 2).
The white radiation has photon radiations that have an energy of around 10.2 eV.
So, they are just sufficient to transmit an electron from n = 1 to n = 2 level.
Hence, the absorption lines are observed only in the Lyman series.

Question 6:

The Balmer series lies in the visible range. Therefore, it was observed and analysed before the other series. The wavelength range of Balmer series is from 364 nm (for n₂ =$\infty$) to 655 nm (for n₂ =3).

Question 7:

Energy of n^th state of hydrogen is given by
$E_{\mathrm{n}} = \frac{-13.6}{n^2} \mathrm{eV}$
Energy of first excited state (n = 2) of hydrogen, E₁ = $\frac{-13.6}{4} \mathrm{eV}$ = -3.4 eV
This relation holds true when the refrence point energy is zero.Usually the refrence point energy is the energy of the atom when the electron is widely separated from the proton.In the given question, the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton so here our refrence point energy is 10 eV. Earlier The energy of first excited state was -3.4 eV when the refrence point had zero energy but now as the refrence point has shifted so The energy of the first excited state will also shift by the corresponding amount.Thus,
E₁^, = -3.4 eV-10 eV = -13.4 eV
We still write E_n = E₁/n², or r_n = a₀n² because these formulas are independent of the refrence point enegy.   

Question 8:

The 'series limit' refers to the 'shortest wavelength' (corresponding to the maximum photon energy).

The frequency of the radiation emitted for transition from n₁ to n₂  is given by
$f=k\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
Here, k is a constant.

For the series limit of Lyman series,
n₁ = 1
n₂ = $\infty$

Frequency, $f_1=k\left(\frac{1}{1^2}-\frac{1}{\infty }\right)=k$

For the first line of Lyman series,
n₁ = 1
n₂ = 2

Frequency, $f_2=k\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3k}{4}$

For series limit of Balmer series,
n₁ = 2
n₂ = $\infty$

$f_1=k\left(\frac{1}{2^2}-\frac{1}{\infty }\right)=\frac{k}{4}$

f₁$-$f₃= f₂

Thus, the difference in the frequencies of series limit of Lyman series and Balmer series is equal to the frequency of the first line of the Lyman series.

Question 9:

The electron volt is the amount of energy given to an electron in order to move it through the electric potential difference of one volt.
1 eV = 1.6 × 10^–19 J
The numerical value of ionisation energy in eV is equal to the ionisation potential in volts. The equality does not hold if these quantities are measured in some other units.

Question 10:

When a photon of energy (E₂ – E₁ = hυ) is incident on an atom in the ground state, the atom in the ground state E₁ may absorb the photon and jump to a higher energy state (E₂). This process is called stimulated absorption or induced absorption.

Spontaneous absorption is the process by which an atom in its ground state spontaneously jumps to a higher energy state, resulting in the absorption of a photon.
We do not have any process such as spontaneous absorption. This is because for absorption, we need to incident a photon of sufficient energy on the atom to stimulate the atom for absorption.

Question 1:

When an atom transits from an excited state to ground state in the presence of an external radiation then it is called as stimulated transition.
When an atom transits from an excited state to ground state on its own then it is called as spontaneous transition.
Ratio of the coefficient for stimulated transition to spontaneous transition is given by
$R = \frac{A}{B\rho \left(\nu \right)} = e^{\frac{h\nu }{kT}}-1$
For microwave region
$$\begin{gathered} \nu = {10}^{10 } \left(\mathrm{say}\right) \\ \mathrm{R} = {\mathrm{e}}^{\frac{6.6\times {10}^{-34}\times {10}^{10}}{1.38\times {10}^{-23}\times 300}}-1 \\ ={\mathrm{e}}^{0.0016}-1 \\ =0.0016 \end{gathered}$$
This implies that stimulated transition dominate in this region.
For visible region
$$\begin{gathered} \nu = {10}^{15} \\ \Rightarrow R = e^{160}-1 \\ \Rightarrow R>>1 \end{gathered}$$
So here spontaneous transition dominate.

Question 2:

(c) h/2π

According to Bohr's atomic theory, the orbital angular momentum of an electron is an integral multiplt of h/2π.
∴  $L_{\mathrm{n}}=\frac{nh}{2\pi }$

Here,
n = Principal quantum number

The minimum value of n is 1.
Thus, the minimum value of the orbital angular momentum of the electron in a hydrogen atom is given by
$L=\frac{h}{2\pi }$

Question 3:

(d) either two atoms or three atoms

The energies of the photons emitted can be expressed as follows:
$13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \mathrm{eV} = 10.2 \mathrm{eV}$
$13.6\left(\frac{1}{1^2}-\frac{1}{3^2}\right) \mathrm{eV} = 12.1 \mathrm{eV}$
$13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \mathrm{eV} = 1.9 \mathrm{eV}$

The following table gives the transition corresponding to the energy of the photon:
 

Question 4:

(c) 10⁻⁴²N-m

The angular momentum of the electron for the nth state is given by
$L_{\mathrm{n}}=\frac{n\mathrm{h}}{2\mathrm{\pi }}$
Angular momentum of the electron for n = 3, $L_{\mathrm{i}}=\frac{3h}{2\mathrm{\pi }}$

Angular momentum of the electron for n = 2, $L_{\mathrm{f}}=\frac{2h}{2\mathrm{\pi }}$

The torque is the time rate of change of the angular momentum.
$$\begin{gathered} \mathrm{Torque}, \tau =\frac{L_{\mathrm{f}}-L_{\mathrm{i}}}{t} \\ =\frac{(2\mathrm{h}/2\mathrm{\pi })-(3\mathrm{h}/2\mathrm{\pi })}{{10}^{-8}} \\ =\frac{-(\mathrm{h}/2\mathrm{\pi })}{{10}^{-8}} \\ =\frac{-{10}^{-34}}{{10}^{-8}} \left[\because \frac{\mathrm{h}}{2\mathrm{\pi }}\approx {10}^{-34} \mathrm{J}-\mathrm{s}\right] \\ =-{10}^{-42} \mathrm{N}-\mathrm{m} \end{gathered}$$
The magnitude of the torque is 10^$-$42 N-m.

Question 5:

(d) n = 2 to n = 1

For the transition in the hydrogen-like atom, the wavelength of the emitted radiation is calculated by

$\frac{1}{\lambda }=\mathrm{R}{Z}^2\left(\frac{1}{n_1}-\frac{1}{n_2}\right)$
Here, R is the Rydberg constant.

For the transition from n = 5 to n = 4, the wavelength is given by
$$\begin{gathered} \frac{1}{\lambda }=\mathrm{R}{Z}^2\left(\frac{1}{4^2}-\frac{1}{5^2}\right) \\ \lambda =\frac{400}{9\mathrm{R}{Z}^2} \end{gathered}$$


For the transition from n = 4 to n = 3, the wavelength is given by
$$\begin{gathered} \frac{1}{\lambda }=\mathrm{R}{Z}^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right) \\ \lambda =\frac{144}{7\mathrm{R}{Z}^2} \end{gathered}$$


For the transition from n = 3 to n = 2, the wavelength is given by
$$\begin{gathered} \frac{1}{\lambda }=\mathrm{R}{Z}^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ \lambda =\frac{36}{5\mathrm{R}{Z}^2} \end{gathered}$$


For the transition from n = 2 to n = 1, the wavelength is given by
$$\begin{gathered} \frac{1}{\lambda }=\mathrm{R}{Z}^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\ \lambda =\frac{2}{\mathrm{R}{Z}^2} \end{gathered}$$

From the above calculations, it can be observed that the wavelength of the radiation emitted for the transition from n = 2 to n = 1 will be minimum.

Question 6:

(d) Doubly ionized lithium

For a hydrogen-like ion with Z protons in the nucleus, the radius of the nth state is given by

$r_{\mathrm{n}}=\frac{n^2{a}_0}{Z}$

Here, a₀ = 0.53 pm

For lithium,
Z = 3
Therefore, the radius of the first orbit for doubly ionised lithium will be minimum.

Question 7:

(d) Doubly ionized lithium

The wavelength corresponding the transition from n₂ to n₁ is given by
$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
Here,
R = Rydberg constant
Z = Atomic number of the ion

From the given formula, it can be observed that the wavelength is inversely proportional to the square of the atomic number.
Therefore, the wavelength corresponding to n = 2 to n = 1 will be minimum in doubly ionized lithium ion because for lithium, Z = 3.

Question 8:

(c)

The speed (v) of electron can be expressed as
$v=\frac{Z{e}^2}{2{\in }_{0}hn}$    ....(1)
Here,
Z = Number of protons in the nucleus
e = Magnitude of charge on electron charge
n = Principal quantum number
h = Planck's constant

It can be observed from equation (1) that the velocity of electron is inversely proportional to the principal quantum number (n).
Therefore, the graph between them must be a rectangular hyperbola.
The correct curve is (c).

Question 9:

(b) increases

The electric potential energy of hydrogen atom with electron at the nth state is given by
$V=-\frac{2\times 13.6}{n^2}$
As the value of n increases, the potential energy of the hydrogen atom also increases, i.e. the atom becomes less bound as n increases.

Question 10:

(c) He⁺

The total energy of a hydrogen-like ion, having Z protons in its nucleus, is given by
$E=-\frac{13.6Z^2}{n^2}$ eV

Here, n = Principal quantum number

For ground state,
n = 1
∴ Total energy, E = $-$ 13.6 Z² eV

For hydrogen,
Z = 1
∴ Total energy, E = $-$ 13.6 eV

For deuterium,
Z = 1
∴ Total energy, E = $-$ 13.6 eV

For He⁺,
Z = 2
 ∴ Total energy, E = $-$ 13.6×2² = $-$ 54.4 eV

For Li⁺⁺,
Z = 3
∴ Total energy, E = $-$ 13.6×3² = $-$ 122.4 eV

Hence, the ion having an energy of −54.4 eV in its ground state may be He⁺.

Question 11:

(d) Li⁺⁺

The radius of the nth orbit in one electron system is given by
$r_{\mathrm{n}}=\frac{n^2{a}_0}{Z}$

Here, a₀ = 53 pm

For the shortest orbit,
n = 1

For hydrogen,
Z = 1

∴ Radius of the first state of hydrogen atom = 53 pm


For deuterium,
Z= 1

∴ Radius of the first state of deuterium atom = 53 pm


For He⁺,
Z = 2

∴ Radius of He⁺ atom = $\frac{53}{2} \mathrm{pm} =26.5 \mathrm{pm}$


For Li⁺⁺,
Z = 3

∴ Radius of Li⁺⁺ atom = $\frac{53}{3} \mathrm{pm} =17.66 \mathrm{pm}\approx 18 \mathrm{pm}$

The given one-electron system having radius of the shortest orbit to be 18 pm may be Li⁺⁺.

Question 12:

(a) 1.05 × 10⁻³⁴ J s

Let after absorption of energy, the hydrogen atom goes to the nth excited state.

Therefore, the energy absorbed can be written as
$$\begin{gathered} 10.2=13.6\times \left(\frac{1}{1^2}-\frac{1}{n^2}\right) \\ \Rightarrow \frac{10.2}{13.6}=1-\frac{1}{n^2} \\ \Rightarrow \frac{1}{n^2}=\frac{13.6-10.2}{13.6} \\ \Rightarrow \frac{1}{n^2}=\frac{3.4}{13.6} \\ \Rightarrow n^2=4 \\ \Rightarrow n=2 \end{gathered}$$

The orbital angular momentum of the electron in the nth state is given by
$L_{\mathrm{n}}=\frac{nh}{2\pi }$
Change in the angular momentum, $∆L=\frac{2h}{2\pi }-\frac{h}{2\pi }=\frac{h}{2\pi }$
∴ $∆L=1.05\times {10}^{-34} \mathrm{Js}$

Question 13:

(d) Orbital angular momentum of the electron

According to Bohr's atomic theory, the orbital angular momentum of an electron in a one-electron system is given by
$L_{\mathrm{n}}=\frac{nh}{2\mathrm{\pi }}$

Here,
n = Principal quantum number

The angular momentum is independent of the atomic number of the one-electron system. Therefore, it is same for all hydrogen-like atoms and ions in their ground states.
The other parameters given here are dependent on the atomic number of the hydrogen-like atom or ion taken.

Question 1:

(d) move with same speed

All the photons emitted in the laser move with the speed equal to the speed of light (c = 3×10⁸ m/s).

Ideally, the light wave through the laser must be coherent, but in practical laser tubes, there is some deviation from the ideal result. Thus, the photons emitted by the laser have little variations in their wavelengths and energies as well as the directions, but the velocity of all the photons remains same.

Question 2:

(b) the temperature of hydrogen is much smaller than that of the star

The number of lines of the hydrogen spectrum depends on the excitation of the hydrogen atom. This is dependent on the heat energy absorbed by the hydrogen atoms. More the temperature of the hydrogen sample, more is the heat energy. The temperature of hydrogen at the star is much more than that can be produced in the laboratory. Hence, less number of lines are observed in the hydrogen spectrum in the laboratory than that in a star.

Question 3:

(a) must be elastic.
The minimum energy required to excite a hydrogen atom from its ground state to 1st excited state is approximately 10 eV. As the incident electron energy is not sufficient for excitation of the hydrogen atom so electron will not get absorbed in the hydrogen atom so it can not be an inelastic collision. Also this collision can not be partially elastic because in an partially elestic collision, there is a net loss on kinetic energy. If the energy is lost then corresponding amount of heat shlould have been produced but it is not so which implies that the collision is completely elastic. 

Question 4:

(a) vn
(b) Er
Relations for energy, radius of the orbit and its velocity are given by
$$\begin{gathered} E = -\frac{\mathrm{m}{Z}^2{\mathrm{e}}^4}{8{{\in }_0}^2{\mathrm{h}}^2{n}^2} \\ r = \frac{{\in }_0{\mathrm{h}}^2{n}^2}{{\mathrm{\pi mZe}}^2} \\ v= \frac{Z{\mathrm{e}}^2}{2{\in }_0\mathrm{h}n} \end{gathered}$$
Where
Z : the atomic number of hydrogen like atom
e : electric charge
h : plank constant
m : mass of electron
n : principal quantam number of the electron
${\in }_0$ : permittivity of vacuum
From these relations, we can see that the products independent of n are  vn, Er.

Question 5:

(a) will pass through the origin
(b) will be a straight line with slope 4

The radius of the nth orbit of a hydrogen atom is given by
$r_{\mathrm{n}}=n^2{a}_0$
Area of the nth orbit is given by
$$\begin{gathered} A_{\mathrm{n}} = \mathrm{\pi }{r}_{\mathrm{n}}^2 = \mathrm{\pi }{n}^4{a}_0^2 \\ {A}_1 = \mathrm{\pi }{a}_0^2 \\ \Rightarrow \ln \left(\frac{A_{\mathrm{n}}}{A_1}\right)= \ln \left(\frac{\mathrm{\pi }{n}^4{a}_0^2}{\mathrm{\pi }{a}_0^2}\right) \\ \ln \left(\frac{A_{\mathrm{n}}}{A_1}\right)= 4\ln n ...\left(1\right) \end{gathered}$$
From the above expression, the graph of ln (A_n/A₁) against ln(n) will be a straight line passing through the origin and having slope 4.

Question 6:

(b) u_A > u_B
The ionisation energy of a hydrogen like ion of atomic number Z is given by
$V=(13.6 \mathrm{eV})\times Z^2$

Thus, the atomic number of ion A is greater than that of B (Z_A > Z_B).
The radius of the orbit is inversely proportional to the atomic number of the ion.
∴ r_A > r_B
Thus, (a) is incorrect.
The speed of electron is directly proportional to the atomic number.
Therefore, the speed of the electron in the orbit of A will be more than that in B.
Thus, u_A > u_Bis correct.
The total energy of the atom is given by
$E=-\frac{m{Z}^2{e}^2}{8{\in }_0{h}^2{n}^2}$
As the energy is directly proportional to Z², the energy of A will be less than that of B, i.e.  E_A < E_B.
The orbital angular momentum of the electron is independent of the atomic number.
Therefore, the relation L_A > L_B is invalid.

Question 1:

(a) same energy
(b) same direction
(c) same phase
(d) same wavelength

When a photon stimulates the emission of another photon, the two photons have same energy, direction, phase, and wavelength or we can say that the two photons are coherent.
When an atom is present in its excited state then if a photon of energy equal to the energy gap between the excited state and any lower stable state is incident on this atom then the atom transits from upper state to the lower stable state by emitting a photon of energy equal to the energy gap between the two states. It is called stimulated emission. The emitted photon and incident photon have same energy and hence same wavelength. Also these two photons will be in phase and in the same direction. This process of producing monochromatic  and unidirectional light is called lasing action.

Question 2:

The dimensions of ε₀ can be derived from the formula given below:

$$\begin{gathered} a=\frac{{\epsilon }_0 h^2}{\mathrm{\pi }m{e}^2}=\frac{{\mathrm{A}}^2{\mathrm{T}}^2{\left({\mathrm{ML}}^2{\mathrm{T}}^{-1}\right)}^2}{{\mathrm{L}}^2{\mathrm{ML}}^{-2}\mathrm{M}{\left(\mathrm{AT}\right)}^2} \\ =\frac{{\mathrm{M}}^2{\mathrm{L}}^2{\mathrm{T}}^{-2}}{{\mathrm{M}}^2{\mathrm{L}}^3{\mathrm{T}}^{-2}}=\mathrm{L} \end{gathered}$$

Clearly, a₀ has the dimensions of length.

Question 3:

From Balmer empirical formula, the wavelength $\left(\lambda \right)$ of the radiation is given by
$\frac{1}{\lambda }=R\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$
Here, R = Rydberg constant = 1.097$\times$10⁷ m^$-1$ 
           n₁ = Quantum number of final state
           n₂ = Quantum number of initial state
(a)
For transition from n = 3 to n = 2:

Here,
n₁ = 2
n₂ = 3

$$\begin{gathered} \frac{1}{\mathrm{\lambda }}=1.09737\times {10}^7\times \left(\frac{1}{4}-\frac{1}{9}\right) \\ \Rightarrow \mathrm{\lambda }=\frac{36}{5\times 1.09737\times {10}^7} \\ =6.56\times {10}^{-7}=656 \mathrm{nm} \end{gathered}$$

(b)
For transition from n = 5 to n = 4:

Here,
n₁ = 4
n₂ = 5

$$\begin{gathered} \frac{1}{\lambda }=1.09737\times {10}^{-7} \left(\frac{1}{16}-\frac{1}{25}\right) \\ \Rightarrow \lambda =\frac{400}{1.09737\times {10}^7\times 9} \\ =4050 \mathrm{nm} \end{gathered}$$

(c)
For transition from n = 10 to n = 9:

Here,
n₁ = 9
n₂ = 10

$$\begin{gathered} \frac{1}{\lambda }=1.09737\times {10}^7 \left(\frac{1}{81}-\frac{1}{100}\right) \\ \lambda =\frac{81\times 100}{19\times 1.09737\times {10}^7} \\ =38849 \mathrm{nm} \end{gathered}$$

Question 4:

Given:
For the smallest wavelength, energy should be maximum.
Thus, for maximum energy, transition should be from infinity to the ground state.

∴ n₁= 1 
n₂= $\infty$ 

(a) Wavelength of the radiation emitted $\left(\lambda \right)$ is given by
$\frac{\mathit{1}}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For hydrogen,
Atomic number, Z = 1
R = Rydberg constant = 1.097×10⁷ m^$-$1

On substituting the respective values,          
$$\begin{gathered} \lambda =\frac{1}{1.097\times {10}^7}=\frac{1}{1.097}\times {10}^{-7} \\ =0.911\times {10}^{-7} \\ =91.1\times {10}^{-9}=91 \mathrm{nm} \end{gathered}$$

(b)
For He⁺,
Atomic number, Z = 2
Wavelength of the radiation emitted by He⁺ ($\lambda$) is given by
$\frac{\mathit{1}}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
$\therefore \frac{1}{\lambda }={\left(2\right)}^2(1.097\times {10}^7) \left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(\infty \right)}^2}\right)$
$\Rightarrow \lambda =\frac{91 \mathrm{nm}}{4}=23 \mathrm{nm}$

(c) For Li⁺⁺,
 Atomic number, Z = 3
Wavelength of the radiation emitted by  Li⁺⁺ ($\lambda$) is given by
$\frac{\mathit{1}}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
$\therefore \frac{1}{\lambda }={\left(3\right)}^2\times (1.097\times {10}^7) \left(\frac{1}{1^2}-\frac{1}{{\infty }^2}\right)$
$\Rightarrow \lambda =\frac{91 \mathrm{nm}}{Z^2}=\frac{91}{9}=10 \mathrm{nm}$

Question 5:

Expression of Rydberg constant (R) is given by
$R=\frac{m{e}^4}{8h^{2}c {\in }_0^2}$
Mass of electron, m_e = 9.31$\times$10²¹ kg
Charge, e = 1.6 × 10⁻¹⁹ C  
Planck's constant, h = 6.63 × 10⁻³⁴ J-s,  
Speed of light, c = 3 × 10⁸ m/s,
Permittivity of vacuum, ∈₀ = 8.85 × 10⁻¹²  C²N^$-1$m
On substituting the values in the expression, we get
^{$$\begin{gathered} R=\frac{\left(9.31\times {10}^{-31}\right)\times {\left(1.6\times {10}^{-19}\right)}^4}{8\times {\left(6.63\times {10}^{-34}\right)}^2\times \left(3\times {10}^8\right)\times {\left(8.85\times {10}^{-12}\right)}^2} \\ \Rightarrow R =1.097\times {10}^7 {\mathrm{m}}^{-1} \end{gathered}$$}

Question 6:

The binding energy (E) of hydrogen atom is given by
$E=\frac{-13.6}{n^2} \mathrm{eV}$
For state n = 2,
$$\begin{gathered} E =-\frac{ 13.6}{{\left(2\right)}^2} \\ \Rightarrow E=-3.4 \mathrm{eV} \end{gathered}$$

Thus, binding energy of hydrogen at n = 2 is $-3.4 \mathrm{eV}$. 

Question 7:

For He⁺ ion,
Atomic number, Z = 2
For hydrogen like ions, radius $\left(r\right)$ of the nth state is given by
$r=\frac{0.53 n^2}{Z}$Å
Here,
Z = Atomic number of ions
n = Quantum number of the state

Energy $\left(E\right)$ of the nth state is given by
E_n = $-\frac{13.6 Z^2}{n^2}$
(a)
For n = 1,
Radius, $$\begin{gathered} r=\frac{0.53\times {\left(1\right)}^2}{2} \\ = 0.265 {\mathrm{A}}^0 \end{gathered}$$ Å

Energy, E_n = $\frac{-13.6\times 4}{1}$
  = $-$ 54.4 eV      

(b)
For n = 4,
Radius, $r=\frac{0.53\times 16}{2}=4.24 \mathrm{\AA }$
Energy, $E=\frac{-13.6\times 4}{16}=-3.4 e\mathrm{V}$
(c)
For n = 10,
Radius, $r=\frac{0.53\times 100}{2}$
    = 26.5 Å
Energy, $E=\frac{-13.6\times 4}{100}=-0.544 \mathrm{eV}$

Question 8:

Wavelength of ultraviolet radiation, $\lambda$ = 102.5 nm = 102,5 $\times$10^$-9$ m
Rydberg's constant, R = 1.097$\times$ 10⁷ m^$-1$
Since the emitted light lies in ultraviolet range, the lines will lie in lyman series.
Lyman series is obtained when an electron jumps to the ground state (n_​1 = 1) from any excited state (n_​2).
Wavelength of light $\left(\lambda \right)$ is given by
$$\begin{gathered} \frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \\ \mathrm{Here}, R=\mathrm{Rydberg} \mathrm{constant} \\ \Rightarrow \frac{1}{102.5\times {10}^{-9}}=1.097\times {10}^7 \left(\frac{1}{1^2}-\frac{1}{n_2^2}\right) \\ \Rightarrow \frac{{10}^9}{102.5}=1.097\times {10}^7 \left(1-\frac{1}{n_2^2}\right) \\ \Rightarrow \frac{{10}^2}{102.5}=1.097 \left(1-\frac{1}{n_2^2}\right) \\ \Rightarrow 1-\frac{1}{n_2^2}=\frac{100}{102.5\times 1.097} \\ \Rightarrow \frac{1}{n_2^2}=1-\frac{100}{102.5\times 1.097} \\ \Rightarrow n_2=\sqrt{9.041}=3 \end{gathered}$$
The transition will be from 1 to 3.

Question 9:

(a) PE of hydrogen like atom in the nth state, V = $=\frac{-13.6Z^2}{n^2}\mathrm{eV}$
Here, Z is the atomic number of that atom.
For the first excitation, the atom has to be excited from n = 1 to n = 2 state.
So, its excitation potential will be equal to the difference in the potential of the atom in n = 1 and in n = 2 states. 
First excitation potential of He⁺
$$\begin{gathered} -13.6Z^2 (1-\frac{1}{2^2}) eV \\ =-10.2 Z^2 eV \end{gathered}$$
⇒10.2 × Z²
 = 10.2 × 4
= 40.8 V

(b) Ionization Potential Li⁺⁺ = 13.6 V × Z²
                                        = 13.6 × 9
                                         = 122.4 V

Question 10:

There will be three wavelengths.
(i) For the transition from (n = 4) to (n = 3) state
(ii) For the transition from (n = 3) to (n = 2) state
(iii) For the transition from (n = 4) to (n = 2) state

Let ($\lambda$₁) be the wavelength when the atom makes transition from (n = 4) state to (n = 2) state._​ 
Here,
n₁ = 2
n₂ = 4
Now, the wavelength ($\lambda$₁)​ will be
$\frac{1}{{\lambda }_1}=R\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$
$$\begin{gathered} R=1.097\times {10}^7 {\mathrm{m}}^{-1} \\ \frac{1}{{\lambda }_1}=1.097\times {10}^7\times \left(\frac{1}{4}-\frac{1}{16}\right) \\ \Rightarrow \frac{1}{{\lambda }_1}=1.097\times {10}^7 \left(\frac{4-1}{16}\right) \\ \Rightarrow \frac{1}{{\lambda }_1}=\frac{1.097\times {10}^7\times 3}{16} \\ \Rightarrow {\lambda }_1=\frac{16\times {10}^{-7}}{3\times 1.097} \\ = 4.8617\times {10}^{-7} \\ = 486.1\times {10}^{-9} \\ =487 \mathrm{nm} \end{gathered}$$               

When an atom makes transition from (n = 4) to (n = 3), the wavelength ($\lambda$2) is given by

$$\begin{gathered} \mathrm{Here} \mathrm{again}, \\ {n}_1=3 \\ {n}_2=4 \\ \frac{1}{{\lambda }_2}=1.097\times {10}^7 \left(\frac{1}{9}-\frac{1}{16}\right) \\ \Rightarrow \frac{1}{{\lambda }_2}=1.097\times {10}^7 \left(\frac{16-9}{144}\right) \\ \Rightarrow \frac{1}{{\lambda }_2}=\frac{1.097\times {10}^7\times 7}{144} \\ \Rightarrow {\lambda }_2=\frac{144}{1.097\times {10}^7\times 7} \\ =1875 \mathrm{nm} \end{gathered}$$

Similarly, wavelength ($\lambda$₃) for the transition from (n = 3) to (n = 2) is given by
When the transition is n₁ = 2 to n₂ = 3:
$$\begin{gathered} \frac{1}{{\lambda }_3}=1.097\times {10}^7\left(\frac{1}{4}-\frac{1}{9}\right) \\ \Rightarrow \frac{1}{{\lambda }_3}=1.097\times {10}^7 \left(\frac{9-4}{36}\right) \\ \Rightarrow \frac{1}{{\lambda }_3}=\frac{1.097\times {10}^7\times 5}{36} \\ \Rightarrow {\lambda }_3=\frac{36\times {10}^{-7}}{1.097\times 5}=656 \mathrm{nm} \end{gathered}$$