Question

In hydrogen atom, if the difference in the energy of the electron in n = 2 and n = 3 orbits is E, the ionization energy of hydrogen atom is

• A. 13.2 E
• B. 7.2 E
• C. 5.6 E
• D. 3.2 E

Answer 1

The correct option is B 7.2 E

Energy difference between n = 2 and n = 3; $E=K(\frac{1}{2^2}-\frac{1}{3^2})=K(\frac{1}{4}-\frac{1}{9})=\frac{5}{36}K$ ....... (i)
Ionization energy of hydrogen atom $n_1=1$ and $n_2=\infty ;E^{{}^{\prime }}=K(\frac{1}{1^2}-\frac{1}{{\infty }^2})=K$ ....... (ii)
From equation (i) and (ii) $E^{{}^{\prime }}=\frac{36}{5}E=7.2E$