Question

The electron energy in a 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 the $n=2$ orbit. What among the following is the longest wavelength (in $cm$) of light that can be used to cause this transmission:

• A. $3.66\times {10}^{-6}cm$
• B. $3.66\times {10}^{-5}cm$
• C. $3.66\times {10}^{-7}cm$
• D. $3.66\times {10}^{-5}cm$

Answer 1

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

The required transition is $n_1=2$ to $n_2=\infty$ and corresponding transition energy is:
$\Delta E$ = $21.7\times {10}^{-12}[\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}]erg$
= $\frac{21.7}{4}\times {10}^{-12}erg$
= $5.425\times {10}^{-12}erg$
The longest wavelength that can cause above transition can be determined as:
$\lambda =\frac{hc}{\Delta E}$
= $\frac{6.625\times {10}^{-34}\times 3\times {10}^8}{5.425\times {10}^{-12}\times {10}^{-7}}$
= $3.66\times {10}^{-7}m$
= $3.66\times {10}^{-5}cm$