Atomic Structure Question 468

Question: The ionization energy of hydrogen atom (in the ground state) is x kJ. The energy required for an electron to jump from 2nd orbit to the $ 3^{rd} $ orbit will be

Options:

A) x/6

B) 5x

C) 7.2 x

D) 5x/36

Show Answer

Answer:

Correct Answer: D

Solution:

  • $ {{(IE)} _{H}}={E _{\infty }}-E_1=-E_1=x $ volt

Put $ E_1=-\frac{K}{n^{2}}=-\frac{K}{1^{2}}=-K $

$ \therefore $ $ K=x $ $ \Delta E=E_3-E_2 $ $ =-K/3^{2}-(-K/2^{2}) $ $ =K\frac{5}{36}=\frac{5x}{36} $



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