Question

Calculate the wavelength of radiation emitted when electron in a hydrogen atom jumps from $n=\infty$ to $n=1$.

Answer 1

The energy of $n^{th}$ level for any atom is determined by the radius and quantum number by:

$E_n=-\frac{Z^2{k}_e^2{e}^4{m}_e}{2{\hslash }^2{n}^2}$

For hydrogen atom the energy varies with shell number as:

$E_n=-13.6\frac{1}{n^2}eV$

Energy energy released from jump of electron from $n=\infty$ to $n=1$ is:

$E=-13.6(\frac{1}{\infty }-(\frac{1}{1^2}\left)\right)eV$

$=13.6eV$

Photon of this energy is given by, $E=\frac{hc}{\lambda }$
$⟹\lambda =\frac{hc}{E}$$=91.17nm$