Question

The emission series of hydrogen atom is given by
$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
where, $R$ is the Rydberg constant. For a transition from $n_2$ to $n_1$, the relative change $\Delta \lambda /\lambda$ in the emission wavelength, if hydrogen is replaced by deuterium (assume that, the mass of proton and neutron are the same and approximately $2000$ times larger than that of electrons) is

• A. $0.025\%$
• B. $0.005\%$
• C. $0.0025\%$
• D. $0.05\%$

Answer 1

Answer: (D)

In Bohr’s theory, it is assumed that nucleus of H-atom is so heavy that it remained fixed at the circular orbit, while electron revolves around it. This assumption is true only when mass of nucleus is infinite compared to mass of electron.
Infact the nuclear mass is finite and in $H$-atom, nucleus is only about $2000$ times $(\sim 1836$ times $)$ heavier than orbiting electron.
So, the assumption that nucleus is fixed is not justified.
The nucleus and electron must be revolving around their common centre of mass and orbit must be elliptical.
Let $m_e$ is mass of electron and $m_n$ is mass of nucleus, then distances of electron and nucleus from centre of mass must be

$r_1=\left(\frac{m_e}{m_n+m_e}\right)r$
and $r_2=\left(\frac{m_n}{m_n+m_e}\right)r$
So, orbital angular momentum of atom
$=L_{\text{election}}+L_{\text{proton}}$
$=m_e{r}_2^2\omega +m_n{r}_1^2\omega$
$=\left(\frac{m_e{m}_n}{m_e+m_n}\right)r^2\omega =\mu r^2\omega$
Where, $\frac{m_e{m}_n}{m_e+m_n}=\mu$=reduced mass.
Now, by Bohr’s postulate, we have
$\mu \omega r^2=\frac{nh}{2\pi }$
$\Rightarrow \mu vr=\frac{nh}{2\pi }$
$\therefore$ Energy of electron in $nth$ orbit of a one electron atom is
$E_n=-\left(\frac{\mu Z^2{e}^4}{8n^2{h}^2{\epsilon }_0^2}\right)$
When electron jumps from $n_2$ to $n_1$ state, radiated wavelength will be
$\frac{1}{\lambda }=\frac{\mu e^4{Z}^2}{8h^{3}c{\epsilon }_0}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
Keeping other parameters same, for a given transition $n_2$ to $n_1$, we can say
$\frac{1}{\lambda }\propto \mu Z^2\Rightarrow \frac{1}{\lambda }=C\mu Z^2[\therefore C=\text{ constant }]$
or $\frac{1}{\lambda }=C\left(\frac{m_e{m}_n}{m_e+m_n}\right)\cdot Z^2$
Now, for $H$-atom $(Z=1,A=1)$,
$\frac{1}{{\lambda }_1}=C\cdot \left(\frac{m_e\times 2000m_e}{2001m_e}\right)\times 1$
$\Rightarrow \frac{1}{{\lambda }_1}=C\cdot \frac{2000}{2001}\cdot m_e$
$\Rightarrow {\lambda }_1=\frac{2001}{2000C\cdot m_e}$ ...(i)
For deuterium atom $(Z=1,A=2)$,
$\frac{1}{{\lambda }_2}=C\cdot \left(\frac{m_e\times 4000m_e}{4001m_e}\right)\times 1$
$\Rightarrow \frac{1}{{\lambda }_2}=C\cdot \frac{4000m_e}{4001}$
$\Rightarrow {\lambda }_2=\frac{4001}{4000C\cdot m_e}$ ...(ii)
Subtraction Eq. (ii) by Eq. (i), we have
$\Rightarrow \left|{\lambda }_2-{\lambda }_1\right|=\left|\frac{4001}{4000C\cdot m_e}-\frac{2001}{2000C\cdot m_e}\right|$
$=\frac{2}{4000C\cdot m_e}$
$\Rightarrow \left|\frac{{\lambda }_2-{\lambda }_1}{{\lambda }_1}\right|=\frac{\left(\frac{2}{4000C\cdot m_e}\right)}{\left(\frac{2001}{2000C\cdot m_e}\right)}=\frac{1}{2001}$
So, \% change in wavelength is
$\frac{\Delta \lambda }{{\lambda }_1}\times 100=\frac{1}{2001}\times 100\approx 0.05\%$