Question

The electron energy in hydrogen atom is given by $E_n=(–2.18\times {10}^{–18})/n^2$ J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Answer 1

Given,
$E_n=-\frac{2.18\times {10}^{-18}}{n^2}J$
Energy required for ionization from n = 2 is given by,
$△E=E_{infty}-E_2=[\left(\frac{-2.18\times {10}^{-18}}{(\infty {)}^2}\right)-\left(\frac{-2.18\times {10}^{-18}}{(2{)}^1}\right)]J=[\frac{2.18\times {10}^{-18}}{4}-0]j=0.545\times {10}^{-18}J$
$△E=0.545\times {10}^{-18}J\lambda =\frac{hc}{△E}$
Here, λ is the longest wavelength causing the transition.
$\lambda =\frac{(6.626\times {10}^{-34})(3\times {10}^8)}{5.45\times {10}^{-19}}=3.647\times {10}^{-7}m$

= $3.647\times {10}^{-5}cm$