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)

$3.7 \times 10^{14} Hz$

Answer 2

$\frac{1}{\lambda} = R \left[\frac{1}{\left(n_{1}\right)^{2}} - \frac{1}{\left(n_{2}\right)^{2}}\right]$
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 $n = \frac{c}{\lambda}$
$= \frac{3\times 10^{8}}{8.25 \times 10^{-7}}$
$n = 3.7 \times 10^{14} Hz$