Question

The frequency of one of the lines in the paschen series of a hydrogen atom is $2.33\times {10}^{14}$ Hz. the quantum number $n_2$ which causes this transition is:

• A. 3
• B. 4
• C. 6
• D. 5

Answer 1

Answer: (D)

5

Frequency of paschen series $=2.33\times {10}^{14}Hz=\nu$

we know,$\nu =\frac{c}{\lambda }\Rightarrow \lambda =\frac{c}{\nu }$

$\lambda =\frac{3\times {10}^8}{2.33\times {10}^{14}}=1.288\times {10}^{-6}m$

for paschen series $,n_1=3$

let $n_2=n$

we know the Rydbergs equation

$\frac{1}{\lambda }={Rz}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$\frac{1}{\lambda }=R{\left(1\right)}^2(\frac{1}{3^2}-\frac{1}{n^2})$

$776666.667=10967800\left(\frac{n^2-9}{9n^2}\right)$

$0.6373\times n^2=n^2-9$

$9=n^2\left(0.363\right)$

$n^2=24.815$

$n\cong 5$

$\Rightarrow$ quantum number $n_2$which causes this transition is 5.