Atomic Structure Question 41
Question: The frequency corresponding to transition $ n=2 $ to $ n=1 $ in hydrogen atom is
Options:
A) $ 15.66\times 10^{10}Hz $
B) $ 24.66\times 10^{14}Hz $
C) $ 30.57\times 10^{14}Hz $
D) $ 40.57\times 10^{24}Hz $
Show Answer
Answer:
Correct Answer: B
Solution:
$ v=\frac{1}{\lambda }=R[ \frac{1}{n_1^{2}}-\frac{1}{n_2^{2}} ]=109678[ \frac{1}{1}-\frac{1}{4} ]=82258.5 $
$ \lambda =1.21567\times {10^{-5}}cm\ \ \ or\ \ \ \lambda =12.1567\times {10^{-6}}cm $
$ =12.1567\times {10^{-8}},m $
$ v=\frac{c}{\lambda }=\frac{3\times 10^{8}}{12.567\times {10^{-8}}}=24.66\times 10^{14}Hz $ .
