The electron in Hydrogen atom in ground state revolves on a circular path whose radius is equal to the Bohr radius $\left(a_n\right)$.

Let the velocity of electron is $v$.

$\therefore \quad$ Number of revolutions per unit time $=\frac{2 \pi a_0}{v}$

The electric current is given by $i=\frac{q}{t}$, if $q$ charge flows in timet. Here, $q=e$ The electric current is given by $i=\frac{2 \pi a_0}{v} e$.

The frequency of any line in a series in the spectrum of hydrogen like atoms corresponding to the transition of electrons from $(n+p)$ level to $n$th level can be expressed as a difference of two terms;

$$v_{m n}=c R Z^2\left[\frac{1}{(n+p)^2}-\frac{1}{n^2}\right]$$

where, $m=n+p,(p=1,2,3, \ldots)$ and $R$ is Rydberg constant.

$\begin{aligned} &\begin{aligned} \text{For}\quad p & \ll n \\ v_{m n} & =c R Z^2\left[\frac{1}{n^2}\left(1+\frac{p}{n}\right)^{-2}-\frac{1}{n^2}\right] \\ v_{m n} & =c R Z^2\left[\frac{1}{n^2}-\frac{2 p}{n^3}-\frac{1}{n^2}\right] \end{aligned}\\ &\text { [By binomial theorem } \left.(1+x)^n=1+n x \text { if }|x|<1\right]\\ &v_{m n}=c R Z^2 \frac{2 p}{n^3} \simeq\left(\frac{2 c R Z^2}{n^3}\right) p \end{aligned}$

Thus, the first few frequencies of light that is emitted when electrons fall to the $n$th level from levels higher than $n$, are approximate harmonic (i.e., in the ratio $1: 2: 3 \ldots$ ) when $n \gg 1$.

$H_\gamma$ in Balmer series corresponds to transition $n=5$ to $n=2$. So, the electron in ground state i.e., from $n=1$ must first be placed in state $n=5$.

Energy required for the transition from $n=2$ to $n=5$ is given by

$$=E_1-E_5=13.6-0.54=13.06 \mathrm{eV}$$

Since, angular momentum is conserved, angular momentum coresponding to Hg photon = change in angular momentum of electron

$$\begin{aligned} & =L_5-L_2=5 h-2 h=3 h=3 \times 1.06 \times 10^{-34} \\ & =3.18 \times 10^{-34} \mathrm{~kg}-\mathrm{m}^2 / \mathrm{s} \end{aligned}$$

The total energy of the electron in the stationary states of the hydrogen atom is given by

$$E_n=-\frac{m e^4}{8 n^2 \varepsilon_0^2 h^2}$$

where signs are as usual and the $m$ that occurs in the Bohr formula is the reduced mass of electron and proton in hydrogen atom.

$$\begin{aligned} \text{By Bohr's model,}\quad h v_{i f} & =E_{n_j}-E_{n_f} \\ \text{On simplifying,}\quad v_{i f} & =\frac{m e^4}{8 \varepsilon_0^2 h^3}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \end{aligned}$$

$$\begin{aligned} \text{Since,}\quad\lambda & \propto \frac{1}{\propto} \\ \text{Thus,}\quad\lambda_{i f} & \propto \frac{1}{\propto}\quad\text{.... (i)} \end{aligned}$$

where $\propto$ is the reduced mass. (here, $\propto$ is used in place of $m$ )

Reduced mass for

$$H=\alpha_H=\frac{m_e}{1+\frac{m_e}{M}} ; m_e\left(1-\frac{m_e}{M}\right)$$

$$\begin{aligned} &\text { Reduced mass for }\\ &\begin{aligned} D & =\alpha_D ; m_e\left(1-\frac{m_e}{2 M}\right) \\ & =m_e\left(1-\frac{m_e}{2 M}\right)\left(1+\frac{m_e}{2 M}\right) \end{aligned} \end{aligned}$$

If for hydrogen deuterium, the wavelength is $\frac{\lambda_H}{\lambda_D}$

$$\begin{aligned} & \frac{\lambda_D}{\lambda_H}=\frac{\alpha_H}{\lambda_D} \simeq\left(1+\frac{m_e}{2 M}\right)^{-1} \simeq\left(1-\frac{1}{2 \times 1840}\right) \quad \text{[From Eq. (i)]}\\ & \lambda_D=\lambda_H \times(0.99973) \end{aligned}$$

On substituting the values, we have Thus, lines are $1217.7 \mathop A\limits^o, 1027.7 \mathop A\limits^o, 974.04 \mathop A\limits^o, 951.143 \mathop A\limits^o$.

The total energy of the electron in the $n$th states of the hydrogen like atom of atomic number $Z$ is given by

$$E_n=-\frac{\alpha Z^2 e^4}{8 \varepsilon_0^2 h^2}\left(\frac{1}{n^2}\right)$$

where signs are as usual and the $\propto$ that occurs in the Bohr formula is the reduced mass of electron and proton.

Let $\alpha_H$ be the reduced mass of hydrogen and $\alpha_D$ that of Deutrium. Then, the frequency of the 1 st Lyman line in hydrogen is $h \nu_H=\frac{\alpha_H e^4}{8 \varepsilon_0^2 h^2}\left(1-\frac{1}{4}\right)=\frac{\alpha_H e^4}{8 \varepsilon_0^2 h^2} \times \frac{3}{4}$

Thus, the wavelength of the transition is $\lambda_H=\frac{3}{4} \frac{\alpha_H e^4}{8 \varepsilon_0^2 h^3 \mathrm{C}}$. The wavelength of the transition for the same line in Deutrium is $\lambda_D=\frac{3}{4} \frac{\alpha_D e^4}{8 \varepsilon_0^2 h^3 C}$.

$\therefore$ $\Delta \lambda=\lambda_D-\lambda_H$

$$\begin{aligned} &\text { Hence, the percentage difference is }\\ &\begin{aligned} 100 \times \frac{\Delta \lambda}{\lambda_H} & =\frac{\lambda_D-\lambda_H}{\lambda_H} \times 100=\frac{\alpha_D-\alpha_H}{\alpha_H} \times 100 \\ & =\frac{\frac{m_e M_D}{\left(m_e+M_D\right)}-\frac{m_e M_H}{\left(m_e-M_H\right)}}{\frac{m_e M_H}{\left(m_e+M_H\right.}} \times 100 \\ & =\left[\left(\frac{m_e+M_H}{m_e+M_D}\right) \frac{M_D}{M_H}-1\right] \times 100 \end{aligned} \end{aligned}$$

Since, $m_e \ll M_H \ll M_D$

$$\begin{aligned} \frac{\Delta \lambda}{\lambda_H} \times 100 & =\left[\frac{M_H}{M_D} \times \frac{M_D}{M_H}\left(\frac{1+\frac{m_e}{M_H}}{1+\frac{m_e}{M_D}}\right)-1\right] \times 100 \\ & =\left[\left(1+\frac{m_{\mathrm{e}}}{M_H}\right)\left(1+\frac{m_{\mathrm{e}}}{M_D}\right)^{-1}-1\right] \times 100 \simeq\left[1+\frac{m_{\mathrm{e}}}{M_H}-\frac{m_{\mathrm{e}}}{M_D}-1\right] \times 100 \end{aligned}$$

[By binomial theorem, $(1+x)^n=1+n x$ is $\left.|x|<1\right]$

$$\begin{aligned} & \approx m_e\left[\frac{1}{M_H}-\frac{1}{M_D}\right] \times 100 \\ & =9.1 \times 10^{-31}\left[\frac{1}{1.6725 \times 10^{-27}}-\frac{1}{3.3374 \times 10^{-27}}\right] \times 100 \\ & =9.1 \times 10^{-4}[0.5979-0.2996] \times 100 \\ & =2.714 \times 10^{-2} \% \end{aligned}$$