Question

Calculate the highest frequency of the emitted photon in the Paschen series of spectral lines of the hydrogen atom

• A. $3.7\times 10^{14}Hz$
• B. $9.1\times 10^{15}Hz$
• C. $10.23\times 10^{14}Hz$
• D. $29.7\times 10^{15}Hz$

Answer 1

Answer: (A)

$\frac{1}{\lambda }=R\left[\frac{1}{{\left(n_1\right)}^2}-\frac{1}{{\left(n_2\right)}^2}\right]$
For Paschen series
$\frac{1}{\lambda }=R\left[\frac{1}{{\left(3\right)}^2}-\frac{1}{{\left(\infty \right)}^2}\right]$
$=R\left[\frac{1}{9}-0\right]=R\frac{1}{9}$
$\frac{1}{\lambda }=\frac{R}{9}$
$\lambda =\frac{9}{R}$
$\lambda =\frac{9}{1.09\times 10^7}$
$=8.25\times 10^{-7}$
Frequency $n=\frac{c}{\lambda }$
$=\frac{3\times 10^8}{8.25\times 10^{-7}}$
$n=3.7\times 10^{14}Hz$