Question

The limiting line of Paschen series of H atom will have a frequency equal to:
$R=1.097\times {10}^7{\text{m}}^{-1}$

• A. $3.65\times {10}^{14}$ Hz
• B. $3.65\times {10}^{16}$ Hz
• C. $4.95\times {10}^{17}$ Hz
• D. $3.63\times {10}^{12}$ Hz

Answer 1

The correct option is A $3.65\times {10}^{14}$ Hz

The limiting line will be the shortest line in the Paschen series. It is observed when an electron drops from an orbit with $n=\infty$ to an orbit with $n=3$.

Wave length of this line is given by:
$\frac{1}{\lambda }=R(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})$

Given:
$R=1.097\times {10}^7{\text{m}}^{-1}$
Substituting the values of n and R we get:
$\frac{1}{\lambda }=1.097\times {10}^7\times (\frac{1}{3^2}-\frac{1}{{\infty }^2})\lambda =8.20\times {10}^{-7}{m}^{-1}$
So c = $\nu$$\lambda$
where $\nu$ = frequency of the transition
$\nu$ = $\frac{3\times {10}^8}{8.20\times {10}^{-7}}$ = $3.65\times {10}^{14}$ Hz