JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Modern Physics

Consider an electron in the

n = 3

orbit of a hydrogen-like atom with atomic number

Z

. At absolute temperature

T

, a neutron having thermal energy

k_{B} T

has the same de Broglie wavelength as that of this electron. If this temperature is given by

T = \frac{Z ^{2} h ^{2}}{α π ^{2} a _{0}^{2} m _{N} k _{B}}

, (where

h

is the Planck's constant,

k_{B}

is the Boltzmann constant,

m_{N}

is the mass of the neutron and

a_{0}

is the first Bohr radius of hydrogen atom) then the value of

α

is ____.

Answer: 72

Solution

For the electron in the

n = 3

orbit:

From Bohr's model:

v = \frac{Z e ^{2}}{2 ϵ _{0} nh}

For

n = 3

v = \frac{Z e ^{2}}{6 ϵ _{0} h}

De Broglie wavelength:

λ_{e l ec t r o n} = \frac{h}{m _{e} v}

For the neutron with thermal energy

\frac{1}{2} m_{N} v_{N}^{2} = k_{B} T

λ_{n e u t r o n} = \frac{h}{2 m _{N} k _{B} T}

Equating wavelengths:

\frac{h}{m _{e} v} = \frac{h}{2 m _{N} k _{B} T}

Solving for T:

T = \frac{m _{e}^{2} v ^{2}}{2 m _{N} k _{B}} = \frac{m _{e}^{2} Z ^{2} e ^{4}}{72 ϵ _{0}^{2} h ^{2} m _{N} k _{B}}

Using

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

a_{0}^{2} = \frac{h ^{4} ϵ _{0}^{2}}{π ^{2} e ^{4} m _{e}^{2}}

Substituting:

T = \frac{h ^{2} Z ^{2}}{72 π ^{2} a _{0}^{2} m _{N} k _{B}}

Therefore,

α = 72

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Numerical answer