Question

The electron energy in $H$ atom is given by $\frac{-21.7\times {10}^{-12}}{n^2}$ erg. The longest wavelength (in cm) of light that can be used to cause the removal of an electron from $n=2$ orbit is:

• A. $3.66\times {10}^{-5}$
• B. $3.88\times {10}^{-5}$
• C. $3.66\times {10}^{-6}$
• D. $3.88\times {10}^{-6}$

Answer 1

The correct option is A $3.66\times {10}^{-5}$

The energy required to remove an electron from the second orbit is the same as the energy released when an electron drops from an infinite orbit to the second one.
Therefore, $n_2=\infty$ and $n_1=2.$

Now, by using the given formula,

$\Delta E=-(\frac{21.7\times {10}^{-12}}{{\infty }^2}-\frac{21.7\times {10}^{-12}}{2^2})$$=0+\frac{21.7\times {10}^{-12}}{4}=5.42\times {10}^{-12}erg$
$\Delta E=hv=\frac{hc}{\lambda }=5.42\times {10}^{-12}erg$
$\lambda =\frac{6.62\times {10}^{-27}\times 3\times {10}^{10}}{5.42\times {10}^{-12}}$ $=3.66\times {10}^{-7}m=3.66\times {10}^{-5}cm.$