NCERT Solutions for Chapter: Atoms, Exercise 4: LA

The first four spectral lines in the Lyman series of a H-atom are $\mathrm{\lambda }=1218 \stackrel{\circ }{\mathrm{A},} 1028 \stackrel{\circ }{\mathrm{A}}, 974.3 \stackrel{\circ }{\mathrm{A}} \mathrm{and} 951.4 \stackrel{\circ }{\mathrm{A}}.$ If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

Deutrium was discovered in $1932$ by Harold Urey by measuring the small change in wavelength for a particular transition in ${}^{1}H$ and ${}^{2}H$. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass $\mu$, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here $\mu =m_{e}M/(m_e+M) where M$ is the nuclear mass and $m_e$ is the electronic mass. Estimate the percentage difference in wavelength for the first line of Lyman series in  ${}^{1}H \mathrm{and}{ }^{2}H$. $(\mathrm{Mass} \mathrm{of}{ }^1\mathrm{H} \mathrm{nucleus} \mathrm{is} 1.6725 \times {10}^{-27} \mathrm{kg}, \mathrm{Mass}{ }^2\mathrm{H} \mathrm{nucleus} \mathrm{is} 3.3374\times {10}^{-27} \mathrm{kg}, \mathrm{Mass} \mathrm{of} \mathrm{electron} =9.109\times {10}^{-31}\mathrm{kg}.)$

If a proton had a radius $R$ and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a $\mathrm{H}$-atom when $R=0.1\mathrm{\AA }$

If a proton had a radius $R$ and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a $\mathrm{H}$-atom when $R=10\mathrm{\AA }$

Use, Ground state radius $r_A=\frac{h^2}{4{\pi }^{\mathit{2}}mK{e}^{\mathit{2}}}=0.53\mathrm{\AA }$ , $P.E.=\frac{e^{\mathit{2}}}{\mathit{4}\pi {\epsilon }_{\mathit{0}}{r}_A}=27.2eV$

In the Auger process an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (This is called an Auger electron). Assuming the nucleus to be massive, calculate the kinetic energy of an $\mathrm{n}=4$ Auger electron emitted by Chromium by absorbing the energy from a $\mathrm{n}=2$ to $\mathrm{n}=1$ transition.

The inverse square law in electrostatics is $\left|\mathrm{F}\right|=\frac{e^2}{\left(4{\mathrm{\pi \epsilon }}_0\right).r^2}$for the force between an electron and a proton. The  $\left(\frac{1}{r}\right)$dependence of $\left|\mathrm{F}\right|$can be understood in quantum theory as being due to the fact that the ‘particle’ of light (photon) is massless. If photons had a mass $m_p$, force would be modified to $\left|\mathrm{F}\right|=\frac{e^2}{\left(4{\mathrm{\pi \epsilon }}_0\right)r^2}\left[\frac{1}{r^2}+\frac{\lambda }{r}\right].exp\left(-\lambda r\right)$ where $\lambda =m_{p}c/\overline{h}$ and $\overline{h}=\frac{h}{2\mathrm{\pi }}$ Estimate the change in the ground state energy of a $\mathrm{H}$-atom if $m_p$ were ${10}^{-6}$ times the mass of an electron.

The Bohr model for the $\mathrm{H}$-atom relies on the Coulomb’s law of electrostatics. Coulomb’s law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge $+q_1, –q_2$ is modified to

$\left|\mathrm{F}\right|=\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\left(4{\mathrm{\pi \epsilon }}_0\right)}\frac{1}{{\mathrm{r}}^2},\mathrm{r}\ge {\mathrm{R}}_0$

$=\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{{R_0}^2}{\left(\frac{R_0}{r}\right)}^{\epsilon },r\le {\mathrm{R}}_0$

Calculate in such a case, the ground state energy of a $\mathrm{H}$-atom, if $\epsilon =0.1, R_0=1\stackrel{\mathrm{o}}{\mathrm{A}}.$