Question

The electron energy in hydrogen atom is given by $E=\frac{(-21.7\times {10}^{-12})}{n^2}$ ergs. Calculate the wavelength of light required to remove an electron completely if it is present in second orbit.

• A. $3.67\times {10}^{-5}cm$
• B. $3.67\times {10}^{-8}cm$
• C. $3.67\times {10}^{-3}cm$
• D. $3.67\times {10}^{-5}cm$

Answer 1

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

Given- n = 2 & $E=\frac{(-21.7\times {10}^{-12})}{n^2}$ ergs.
$\Delta E=\frac{2.17\times {10}^{-18}}{n^2}J=\frac{2.17\times {10}^{-18}}{4}J=5.42\times {10}^{-19}J$ or

$\because \Delta E=\frac{hc}{\lambda }$
$\therefore \lambda =\frac{hc}{\Delta E}$
$\lambda =\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{5.42\times {10}^{-19}}=3.67\times {10}^{-7}m$
or $3.67\times {10}^{-5}cm$