Question

The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from $n=2$ to $n=1$ state is:

• A. $194.8\text{nm}$
• B. $490.7\text{nm}$
• C. $913.3\text{nm}$
• D. $121.8\text{nm}$

Answer 1

The correct option is D $121.8\text{nm}$

$\Delta E=10.2\text{eV}$ is the energy difference between the state $n=2$ and $n=1$.

$\Delta E=–3.4–(–13.6)=10.2\text{eV}$

$\frac{hc}{\lambda }=10.2\text{eV}$

$\lambda =\frac{hc}{10.2e}\text{m}$
where, $e=1.6\times {10}^{–19}\text{J/V}$

$\lambda =\frac{12400}{10.2}\text{Å}$
(because $hc=12400\text{eV.}{A}^{\circ }$)
$\lambda \simeq 121.8\text{nm}$