Question

The electron energy in hydrogen atom is given by $E=-\frac{21.7\times {10}^{-12}}{n^2}erg$. Calculate the energy required to remove an electron completely from $n=2$ orbit. What is the longest wavelength (in $cm$) of light that be used to cause this transition?

Answer 1

$E=-\frac{21.7\times {10}^{-12}}{n^2}erg$
Electron energy in the $2$nd orbit, i.e., $n=2$,
$E_2=-\frac{21.7\times {10}^{-12}}{2^2}erg=-5.425\times {10}^{-12}erg$
and $E_{\infty }=0$
$\Delta E=$ Change in energy $=E_{\infty }-E_2=5.425\times {10}^{-12}erg$.
Thus, energy required to remove an electron from $2$nd orbit $=5.425\times {10}^{-12}erg$
According to quantum equation,
$\Delta E=h\cdot \frac{c}{\lambda }$
or $\lambda =\frac{hc}{\Delta E}$
$(h=6.625\times {10}^{-27}erg-sec;c=3\times {10}^{10}cm/sec)$
and $\Delta E=5.425\times {10}^{-12}erg$
So, $\lambda =\frac{(6.625\times {10}^{-27})\times (3\times {10}^{10})}{5.425\times {10}^{-12}}$
$=3.7\times {10}^{-5}cm$
Thus, the longest wavelength of light that can cause this transition is $3.7\times {10}^{-5}cm$.