An electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom. Given the Rydberg's constant R = 105 cm-1. The frequency in Hz of the emitted radiation will be :
Answer & explanation
Correct answer: option 3
$\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3 R}{16} \Rightarrow \lambda=\frac{16}{3 R}=\frac{16}{3} \times 10^{-5} cm$
Frequency $n=\frac{c}{\lambda}=\frac{3 \times 10^{10}}{\frac{16}{3} \times 10^{-5}}=\frac{9}{16} \times 10^{15} Hz$