JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Atomic Structure

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [

a_{0}

is Bohr radius]

Answer: (C)

Solution

From Bohr's postulate,

m v r = \frac{nh}{2 π}

, so

v = \frac{nh}{2 π m r}

. The kinetic energy is

K E = \frac{1}{2} m v^{2} = \frac{1}{2} m (\frac{nh}{2 π m r})^{2} = \frac{n ^{2} h ^{2}}{8 π ^{2} m r ^{2}}

. For the second Bohr orbit (

n = 2

), the radius is

r = \frac{a _{0} n ^{2}}{Z} = \frac{a _{0} \times 4}{1} = 4 a_{0}

. Substituting:

K E = \frac{4 h ^{2}}{8 π ^{2} m ( 4 a _{0} ) ^{2}} = \frac{4 h ^{2}}{8 π ^{2} m \times 16 a _{0}^{2}} = \frac{h ^{2}}{32 π ^{2} m a _{0}^{2}}

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Single correct option