Atoms And Nuclei Question 94

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 [EAMCET (Med.) 2000]

Options:

A) 13.2 E

B) 7.2 E

C) 5.6 E

D) 3.2 E

Show Answer

Answer:

Correct Answer: B

Solution:

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



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