Question

In hydrogen atom, if the difference in the energy of the electron in $n=2$ and $n=3$ orbits is $E.$, then ionization energy of hydrogen atom is $3.6nE.$ The value of $n$ is (integer only)

Answer 1

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 $\left(1\right)$,

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

$\therefore n=2$