Atomic Structure Question 48
Question: The frequency of radiation emitted when the electron falls from $ n=4 $ to $ n=1 $ in a hydrogen atom will be (Given ionization energy of H $ =2.18\times {10^{-18}}J,ato{m^{-1}} $ and $ h=6.625\times {10^{-34}}Js $ ) [CBSE PMT 2004]
Options:
A) $ 3.08\times 10^{15}{s^{-1}} $
B) $ 2.00\times 10^{15}{s^{-1}} $
C) $ 1.54\times 10^{15}{s^{-1}} $
D) $ 1.03\times 10^{15}{s^{-1}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ {E_{ionisation}}={E_{\infty }}-E_{n}=\frac{13.6Z_eff^{2}}{n^{2}}eV $ = $ [ \frac{13.6Z^{2}}{n_2^{2}}-\frac{13.6Z^{2}}{n_1^{2}} ] $
$ E=h\nu =\frac{13.6\times 1^{2}}{{{(1)}^{2}}}-\frac{13.6\times 1^{2}}{{{(4)}^{2}}} $ ; $ h\nu =13.6-0.85 $
$ \because $ $ h=6.625\times {10^{-34}} $
$ \nu =\frac{13.6-0.85}{6.625\times {10^{-34}}}\times 1.6\times {10^{-19}} $ = $ 3.08\times 10^{15}{s^{-1}} $ .
