Question

If the ionization energy of hydrogen is $313.8Kcal/mol$, then the energy of electron in $2^{nd}$ excited state will be:

• A. $-113.2K/mole$
• B. $-78.45K/mole$
• C. $-313.8/mole$
• D. none of these

Answer 1

The correct option is D none of these

$2^{nd}$ excited state means $n=3$

Energy varies inversely with square of number of excited state, i.e.,

$E_n=-\frac{E_1}{n^2}$

$\therefore E_2=-\frac{313.8}{9}=-34.87Kcalmo{l}^{-1}$

Option D is correct.