Question

An energy of $68.0eV$ is required to excite a hydrogen-like atom in its second Bohr energy level to third energy level the charge of nucleus is $Ze$. The wavelength of a radiation required to eject the electron from first orbit to infinity is

• A. $2.2nm$
• B. $2.85nm$
• C. $3.2nm$
• D. $2.5nm$

Answer 1

The correct option is D $2.5nm$

Given, $n_1=2$, $n_2=3$ and $\Delta E=68eV$
The difference in energies of two orbits
$\Delta E=13.6Z^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
where $n_1<n_2$
$68=13.6Z^2[\frac{1}{2^2}-\frac{1}{3^2}]$
$=13.6Z^2[\frac{1}{4}-\frac{1}{9}]$
$=13.6Z^2\left[\frac{9-4}{36}\right]$
$=13.6Z^2\times \frac{5}{36}$
$\therefore Z^2=\frac{68\times 36}{13.6\times 5}=36$
$Z=6$
$\because$ Wavelength of photon, $\frac{1}{\lambda }=Z^{2}R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$n_1=1$ and $n_2=\infty$
$\frac{1}{\lambda }=6^2\times R[\frac{1}{1^2}-\frac{1}{\infty }]=36R$
$\therefore \lambda =\frac{1}{36R}=\frac{1}{36\times 1.097\times {10}^7}$
$=\frac{{10}^{-7}}{39.5}m$
$=0.025\times {10}^{-7}m$
$=2.5nm$