Question

In a hydrogen atom, an electron jumps from the third orbit to the first orbit. Find out the frequency of the spectral line. $(R_H=1.09678\times {10}^7{m}^{-1})$.

• A. $2.925\times {10}^{15}Hz$
• B. $2.925\times {10}^{14}Hz$
• C. $5.85\times {10}^{14}Hz$
• D. $5.85\times {10}^{15}Hz$

Answer 1

The correct option is A $2.925\times {10}^{15}Hz$

$\frac{1}{\lambda }=R_H[\frac{1}{1^2}-\frac{1}{3^2}]$

$C=v\lambda$

$v=\frac{R_H\times C\times 8}{9}=\frac{10967800\times 3\times {10}^8\times 8}{9}$

Frequency$=v=2.925\times {10}^{15}$ $Hz$.