Question

Which electronic transition in a hydrogen atom, starting from the orbit $n=7$, will produce infrared light of wavelength 2170 nm?
(Given: $R_H=1.09677\times {10}^7{m}^{-1}$)

• A. $n=7\to n=6$
• B. $n=7\to n=5$
• C. $n=7\to n=4$
• D. $n=7\to n=3$

Answer 1

The correct option is C $n=7\to n=4$

For H-atom,
$\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
where $R_H=\text{Rydberg constant}$
$n_1=\text{Lower energy state}$
$n_2=\text{Higher energy state}$
Given: $\lambda =2170nm=2170\times {10}^{-9}m$
$R_H=1.09677\times {10}^7{m}^{-1}$

$\therefore \frac{1}{2170\times {10}^{-9}}=1.09677\times {10}^7[\frac{1}{n_1^2}-\frac{1}{7^2}]$

$\Rightarrow n_1=4$

Hence transition $n=7\to n=4$ results into production of infrared light of wavelength 2170 nm