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
Answer & explanation
Correct answer: option 2
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 $E_1=K[1]=\frac{36}{5} E=7.2 E$
∴ lonization energy = 7.2 E