JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Atomic Structure

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

\frac{h ^{2}}{x m a _{0}^{2}}

. The value of

10 x

is ______. (

a_{0}

is radius of Bohr's orbit) (Nearest integer)
[Given :

π = 3.14

]

Answer: 3155

Solution

The kinetic energy of the electron in the

n

th Bohr orbit is

K E = \frac{m e ^{4}}{8 ε _{0}^{2} h ^{2} n ^{2}}

, so for

n = 2

K E = \frac{m e ^{4}}{32 ε _{0}^{2} h ^{2}}

. Using

a_{0} = \frac{ε _{0} h ^{2}}{π m e ^{2}}

, so that

e^{2} = \frac{ε _{0} h ^{2}}{π m a _{0}}

, substitution gives

K E = \frac{h ^{2}}{32 π ^{2} m a _{0}^{2}}

. Comparing with

\frac{h ^{2}}{x m a _{0}^{2}}

gives

x = 32 π^{2} = 32 \times (3.14)^{2} = 315.5

, so

10 x = 3155

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Integer answer