Atomic Structure Question 347
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} $
BETA
