Question

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

• A. $3.67\times {10}^{-7}$
• B. $3.67\times {10}^{-5}$
• C. $3.80\times {10}^{-5}$
• D. $3.80\times {10}^{-7}$

Answer 1

The correct option is B $3.67\times {10}^{-5}$

Energy of an electron in H-atom in $n^{th}$ energy level $\left(E_n\right)=-\frac{21.7\times {10}^{-12}}{n^2}$
Energy required for removing the electron from $n^{th}$ energy level $\left(\Delta E\right)$ = $0-(-\frac{21.7\times {10}^{-12}}{n^2})$
= $\frac{21.7\times {10}^{-12}}{n^2}$
Energy required to remove the electron from $2^{nd}$ orbit for H-atom
$=\frac{21.7\times {10}^{-12}}{4}=5.42\times {10}^{-12}$ erg
Also
$\Delta E=\frac{hc}{\lambda }$
$\lambda =\frac{6.626\times {10}^{-27}\times 3\times {10}^8}{5.42\times {10}^{-12}}$
$=3.67\times {10}^{-7}m$
or $3.67\times {10}^{-5}$ cm