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JEE Advanced 2025Paper 1
Exam
JEE Advanced
Year
2025
Paper
Paper 1
Subject
Physics
Chapter
Atomic Physics

JEE Advanced 2025 · Paper 1 · Physics · Atomic Physics

Consider an electron in the n=3 orbit of a hydrogen-like atom with...

Consider an electron in the n=3n=3 orbit of a hydrogen-like atom with atomic number ZZ. At absolute temperature TT, a neutron having thermal energy kBTk_{\mathrm{B}} T has the same de Broglie wavelength as that of this electron. If this temperature is given by T=Z2h2απ2a02mNkBT=\frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}, (where hh is the Planck's constant, kBk_B is the Boltzmann constant, mNm_{\mathrm{N}} is the mass of the neutron and a0a_0 is the first Bohr radius of hydrogen atom) then the value of α\alpha is ________

Correct answer

0

Step-by-step solution

mv2r=KZe2r2mv2r=14πϵ0Ze2mvr=nh2π\frac{\mathrm{mv}^2}{\mathrm{r}}=\frac{\mathrm{KZe}^2}{\mathrm{r}^2} \mathrm{mv}^2 \mathrm{r}=\frac{1}{4 \pi \epsilon_0} \mathrm{Ze}^2 \mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} (1)/(2) gives v=Ze24πϵ0nh2π=Ze22ϵ0nh\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}}hmv=h2mN⋅KBTT=m2Z2e48ϵ02n2h2mNKBn=3⇒T=m2Z2e472ϵ02h2mNKB(1)(2)2⇒1mr=Ze24πϵ0n2h24π2\begin{aligned} & \frac{h}{m v}=\frac{h}{\sqrt{2 m_N \cdot K_B T}} \\ & T=\frac{m^2 Z^2 e^4}{8 \epsilon_0^2 n^2 h^2 m_N K_B} \\ & n=3 \Rightarrow T=\frac{m^2 Z^2 e^4}{72 \epsilon_0^2 h^2 m_N K_B} \\ & \frac{(1)}{(2)^2} \Rightarrow \frac{1}{m r}=\frac{\frac{Z e^2}{4 \pi \epsilon_0}}{\frac{n^2 h^2}{4 \pi^2}}\end{aligned}r=n2 h2ϵ0πZe2⋅ m⇒a0=h2ϵ0πe2 ma02=h4ϵ02π2e4 m2Ta02=m2Z2e472ϵ0 h2 mNkB⋅h4ϵ02π2e4 m2 T=h2Z272π2a02 mNkB⇒α=72\begin{aligned} & \mathrm{r}=\frac{\mathrm{n}^2 \mathrm{~h}^2 \epsilon_0}{\pi \mathrm{Ze}^2 \cdot \mathrm{~m}} \Rightarrow \mathrm{a}_0=\frac{\mathrm{h}^2 \epsilon_0}{\pi \mathrm{e}^2 \mathrm{~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\end{aligned}$$
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