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

The correct option is B $7.2E$

The energy of the $n^{th}$ state of a hydrogen atom is given by,

$E=-\frac{13.6\text{eV}}{n^2}$

The difference in energy between $n=2$ and $n=3$ is given by,

$\Delta E=E=E_3-E_2$

$\Rightarrow E=13.6\text{eV}[\frac{1}{2^2}-\frac{1}{3^2}]$

$\Rightarrow E=13.6\text{eV}\left[\frac{5}{36}\right]...\left(1\right)$

The ionization energy is defined as the energy required to remove an electron from $n=1$ to $n_2=\infty .$

$|E_{ion}|=|E_{\infty }-E_1|=|0-E_1|$

$\Rightarrow E_{ion}=13.6\text{eV}...\left(2\right)$

Substituting $\left(2\right)$ in $\left(1\right)$,

$E_{ion}=\frac{36}{5}E=7.2E$

Hence, option $\left(B\right)$ is correct.

Why this question? Key concept: The ionization energy is the energy required to remove an electron from $n_1=n$ to $n_2=\infty .$ $E_{ion}=E_{\infty }-E_n=-E_n$