Question

The energy emitted by the electrons when 1g of hydrogen atom undergo transition, giving the spectral lines of lowest energy in the visible region of atomic spectra is $n\times {10}^{-19}J$. The value of n (nearest integer) is :

Answer 1

Use,
$\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

For lowest transition in visible region, transition happens from $n=3$ to $n=2$.
$=1.1\times {10}^7[\frac{1}{2^2}-\frac{1}{3^2}]$
$=1.1\times {10}^7\times \frac{5}{36}$

$\lambda =6.55\times {10}^{-7}$metre

$E=\frac{hc}{\lambda }=\frac{6.62\times {10}^{-34}\times 3.0\times {10}^8}{6.55\times {10}^{-7}}$
$=3.03\times {10}^{-19}J$

$\therefore n=3$

Note: Visible spectral lines occur only in the Balmer series of the hydrogen atom.