Question

A single electron atom has nuclear charge $+Ze$, where $Z$ is atomic number and $e$ is electronic charge. It requires $42.7eV$ to excite the electron from the second Bohr's orbit to third Bohr's orbit.

Answer 1

(a) $\because 1eV=1.602\times {10}^{-12}erg$
Also, $\Delta E=\frac{hc}{\lambda }=E_3-E_2=R_H\times c\times h\times Z^2\left[\frac{1}{2^2}-\frac{1}{3^2}\right]$
or $42.7\times 1.602\times {10}^{-12}=109678\times Z^2\times \left[\frac{1}{2^2}-\frac{1}{3^2}\right]\times 3\times {10}^{10}\times 6.626\times {10}^{-27}$
$Z^2=22.6$ $\therefore Z=5\left(integer\right)$
(b)$\Delta E=E_4-E_3=R_H\times c\times h\times Z^2[\frac{1}{3^2}-\frac{1}{4^2}]$
$=109678\times 3\times {10}^{10}\times 6.626\times {10}^{-27}\times 5^2\times \frac{7}{16\times 9}$
$=26.5\times {10}^{-12}erg$
(c) $\frac{1}{\lambda }=R_H\times Z^2\left[\frac{1}{1^2}-\frac{1}{{\infty }^2}\right]$
$\frac{1}{\lambda }=109678\times 25$
$\therefore \lambda =3.65\times {10}^{-7}cm$
(d)$KE=\frac{1}{2}m{u}^2=\frac{1}{2}m{\left(\frac{2\pi Z{e}^2}{nh}\right)}^2=\frac{2{\pi }^2{Z}^2{e}^{4}m}{n^2{h}^2}$
$=\frac{2\times (3.14{)}^2\times 5^2\times (4.803\times {10}^{-10}{)}^4\times 9.108\times {10}^{-28}}{1^2\times (6.626\times {10}^{-27}{)}^2}$
$=5.45\times {10}^{-10}erg$