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

Answer: (B)

7.2 E

Answer 2

Energy difference between n = 2 and n = 3;

$E = K\left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right) = K\left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5}{36}K$ …..(i)

Ionization energy of hydrogen atom $n_{1} = 1$ and $n_{2} = \infty$;

$E^{'} = K\left( \frac{1}{1^{2}} - \frac{1}{\infty^{2}} \right) = K$ …..(ii)

From equation (i) and (ii) $E^{'} = \frac{36}{5}E = 7.2E$.