Question

Calculate the energy of the radiation corresponding to the spectral line of the lowest frequency in Lyman series in the spectrum of H atom. $(R_H=1.09678\times {10}^7{m}^{-1})$

• A. $4.78\times {10}^{-18}J$
• B. $2.97\times {10}^{-19}J$
• C. $16.36\times {10}^{-19}J$
• D. $1.63\times {10}^{-19}J$

Answer 1

The correct option is C $16.36\times {10}^{-19}J$

According to Lyman series,
$\tilde{\nu }$ = $\frac{1}{\lambda }=R_H(\frac{1}{1^2}-\frac{1}{n_2^2})$
where $\tilde{\nu }$ is the wave number of the transition.
As frequency $\nu \propto \tilde{\nu }$, the lowest frequency corresponds to the lowest possible $n_2$. Hence $n_2=2$.
$\therefore$ $\frac{1}{\lambda }=R_H(\frac{1}{1^2}-\frac{1}{2^2})=\frac{3}{4}{R}_H$
$\therefore \lambda$ = $1.216\times {10}^{-7}m$

Now, we know
Energy = $h\nu =h\left(\frac{c}{\lambda }\right)$
$=6.63\times {10}^{-34}Js\times \frac{3\times {10}^{8}m{s}^{-1}}{1.216\times {10}^{-7}m}=16.36\times {10}^{-19}J$