Question

The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is:

• A. 194.8 nm
• B. 490.7 nm
• C. 913.3 nm
• D. 121.8 nm

Answer 1

The correct option is D

121.8 nm

Step 1: Given Data:-

$$\begin{gathered} n_1=1 \\ {n}_2=2 \\ Z=1 \end{gathered}$$

Hydrogen has an atomic number 1.

Step 2: Formula used:-

${\displaystyle \frac{1}{\lambda }}=R{Z}^2[{\displaystyle \frac{1}{{n_i}^2}}-{\displaystyle \frac{1}{{n_h}^2}}]$ where
$\lambda$ is wavelength of the wave
$R$ is Rydberg constant i.e. $1.097\times {10}^7{m}^{-1}$
$Z$ is atomic number of the element
$n_i$ is lower energy level and $n_h$ is higher energy level.

Step 3: Calculating the wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state,

${\displaystyle \frac{1}{\lambda }}=R{Z}^2[{\displaystyle \frac{1}{{n_i}^2}}-{\displaystyle \frac{1}{{n_h}^2}}]$

$$\begin{gathered} \frac{1}{\lambda }=R{Z}^2[\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}] \\ \frac{1}{\lambda }=R\times {\left(1\right)}^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \\ \frac{1}{\lambda }=\frac{3R}{4} \\ \lambda =\frac{4}{3R}R=1.097\times {10}^7{m}^{-1} \\ \lambda =121.8nm \end{gathered}$$

Hence, option D is the correct answer.