JEE Advanced 2025 — Paper 1 — Question 11

$\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\frac{\mathrm{KZe}^{2}}{\mathrm{r}^{2}}$

$m v^{2} r=\frac{1}{4 \pi \epsilon_{0}} Z e^{2}$

$\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$

(1)/(2) gives $\mathrm{v}=\frac{\frac{\mathrm{Ze}^{2}}{4 \pi \epsilon_{0}}}{\frac{\mathrm{nh}}{2 \pi}}=\frac{\mathrm{Ze}^{2}}{2 \epsilon_{0} \mathrm{nh}}$

$\frac{h}{m v}=\frac{h}{\sqrt{2 m_{N} \cdot K_{B} T}}$

$\mathrm{T}=\frac{\mathrm{m}^{2} \mathrm{Z}^{2} \mathrm{e}^{4}}{8 \epsilon_{0}^{2} \mathrm{n}^{2} \mathrm{~h}^{2} \mathrm{~m}_{\mathrm{N}} \mathrm{K}_{\mathrm{B}}}$ $\mathrm{n}=3 \Rightarrow \mathrm{~T}=\frac{\mathrm{m}^{2} \mathrm{Z}^{2} \mathrm{e}^{4}}{72 \epsilon_{0}^{2} \mathrm{~h}^{2} \mathrm{~m}_{\mathrm{N}} \mathrm{K}_{\mathrm{B}}}$

$\frac{(1)}{(2)^{2}} \Rightarrow \frac{1}{\mathrm{mr}}=\frac{\frac{\mathrm{Ze}^{2}}{4 \pi \epsilon_{0}}}{\frac{\mathrm{n}^{2} \mathrm{~h}^{2}}{4 \pi^{2}}}$

$r=\frac{n^{2} h^{2} \epsilon_{0}}{\pi Z e^{2} \cdot m} \Rightarrow a_{0}=\frac{h^{2} \epsilon_{0}}{\pi e^{2} m}$

$\mathrm{a}_{0}^{2}=\frac{\mathrm{h}^{4} \epsilon_{0}^{2}}{\pi^{2} \mathrm{e}^{4} \mathrm{~m}^{2}}$

$\mathrm{Ta}_{0}^{2}=\frac{\mathrm{m}^{2} \mathrm{Z}^{2} \mathrm{e}^{4}}{72 \epsilon_{0} \mathrm{~h}^{2} \mathrm{~m}_{\mathrm{N}} \mathrm{k}_{\mathrm{B}}} \cdot \frac{\mathrm{h}^{4} \epsilon_{0}^{2}}{\pi^{2} \mathrm{e}^{4} \mathrm{~m}^{2}}$

$\mathrm{T}=\frac{\mathrm{h}^{2} \mathrm{Z}^{2}}{72 \pi^{2} \mathrm{a}_{0}{ }^{2} \mathrm{~m}_{\mathrm{N}} \mathrm{k}_{\mathrm{B}}} \Rightarrow \alpha=72$