Questions
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
a
13.2 E
b
7.2 E
c
5.6 E
d
3.2 E
detailed solution
Correct option is B
Energy, E=K1n12−1n22(K= constant ) n1=2 and n2=3 . So , E=K122−132=K536 For removing an electron n1=1 to n2=∞ Energy E1=K[1]=365E=7.2E ∴ Ionization energy =7.2ETalk to our academic expert!
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