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Question: The ratio of ionization energy of Bohr’s hydrogen atom and Bohr’s hydrogen like Lithium atom is...

The ratio of ionization energy of Bohr’s hydrogen atom and Bohr’s hydrogen like Lithium atom is

A

1:1

B

1:3

C

1: 9

D

None of these

Answer

1: 9

Explanation

Solution

Energy of an electron in ground state of an atom (Bohr’s hydrogen atom like) is given as

E = - 13.6 Z2 eV where Z = atomic number of the atom

⇒ The ionization energy of that atom = Eion = 13.6 Z2

⇒ (Eion)H/Eion)Li = (ZH)2(ZLi)2\frac{(Z_{H})^{2}}{(Z_{Li})^{2}}

⇒ (Eion)H(Eion)Li 6mu= 6mu(13)2= 6mu19\frac{(E_{ion})_{H}}{(E_{ion})_{Li}}\mspace{6mu} = \mspace{6mu}\left( \frac{1}{3} \right)^{2} = \mspace{6mu}\frac{1}{9}