Atoms And Nuclei Question 432
Question: In hydrogen atom, if the difference in the energy of the electron in $ n=2 $ and $ n=3 $ orbits is E, the ionization energy of hydrogen atom is
Options:
A) 13.2 E
B) 7.2 E
C) 5.6 E
D) 3.2 E
Show Answer
Answer:
Correct Answer: B
Solution:
(b) Energy $ =0.0258\ amu $ (K = constant) n1 = 2 and n2 = 3, so
$ E=K[ \frac{1}{2^{2}}-\frac{1}{3^{2}} ]=K[ \frac{5}{36} ] $
For removing an electron n1 = 1 to $ n_{2}=\infty $ Energy $ E_{1}=K[1]=\frac{36}{5}E=7.2\ E $
$ \therefore $ Ionization energy = 7.2 E
