JEE Advanced 2025 — Paper 1 — PHY-III solution

JEE Advanced2025Paper 1PHY-III

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 ___

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Solution

Electron speed in the $n^{th}$ Bohr orbit of a hydrogen-like atom: $v_{e} = \frac{Z e ^{2}}{2 ϵ _{0} nh}$ .

Electron de Broglie wavelength: $λ_{e} = \frac{h}{m _{e} v _{e}}$ .

Neutron de Broglie wavelength at thermal energy $k_{B} T$ : $λ_{n} = \frac{h}{2 m _{N} k _{B} T}$ .

Setting $λ_{e} = λ_{n}$ and squaring:

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

Substituting $v_{e}$ for $n = 3$ :

$T = \frac{m _{e}^{2} Z ^{2} e ^{4}}{2 m _{N} k _{B} \cdot 4 ϵ _{0}^{2} \cdot 9 h ^{2}} = \frac{m _{e}^{2} Z ^{2} e ^{4}}{72 ϵ _{0}^{2} h ^{2} m _{N} k _{B}}$

Now use the Bohr radius identity $a_{0} = \frac{ϵ _{0} h ^{2}}{π m _{e} e ^{2}}$ , so $a_{0}^{2} = \frac{ϵ _{0}^{2} h ^{4}}{π ^{2} m _{e}^{2} e ^{4}}$ . Multiplying numerator and denominator of the expression for $T$ by $\frac{ϵ _{0}^{2} h ^{2}}{π ^{2} m _{e}^{2} e ^{4}}$ :

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

Comparing with the given form: $α = 72$ , i.e. $72.00$ .

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