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

Answer 1

Answer: (B)

Energy, $E=K\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right](K=$ constant $)$
$n_1=2$ and $n_2=3$ So $E=K\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=K\left[\frac{5}{36}\right]$
For removing an electron $n_1=1$ to $n_2=\infty$
Energy $E_1=K[1]=\frac{36}{5}E=7.2E$
$\therefore$ Ionization energy $=7.2E$