Question

Applying Bohr's model when $H$-atom comes from $n=4$ to $n=2$, calculate its wavelength. In this process, write whether energy is released or absorbed? Also write the range of radiation. $R_H=2.18\times {10}^{-18}J,h=6.63\times {10}^{-34}J.s$.

Answer 1

Energy is released in this process; and the radiation will belong to visible region (Balmer series).
$E=\frac{hc}{\lambda }=R_H{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$\frac{1}{\lambda }=\frac{R_H{Z}^2}{hc}[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$=\frac{2.18\times {10}^{-18}\times 1^2}{6.63\times {10}^{-34}\times 3\times {10}^8}[\frac{1}{4}-\frac{1}{16}]$
$=\frac{2.18\times {10}^{-18}\times 1^2}{6.63\times {10}^{-34}\times 3\times {10}^8}\left[\frac{3}{16}\right]$
$\lambda =\frac{6.63\times {10}^{-34}\times 3\times {10}^8\times 16}{3\times 2.18\times {10}^{-18}}=4866\times {10}^{-10}m$
$=4866\mathring{A}$.