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QJEEM24Ph72bc: JEE Main 2024 Physics - Class 12 JEEM24Ph72bc Solution: A hydrogen atom changes its state from to . Due to recoil, the…

A hydrogen atom changes its state from n=3n=3 to n=2n=2 . Due to recoil, the percentage change in the wave length of emitted light is approximately 1×10−n1 \times 10^{-n} . The value of nn is _______.

[Given Rhc =13.6 eV,hc=1242 eV nm,h=6.6×10−34 J s=13.6 \mathrm{~eV}, \mathrm{hc}=1242 \mathrm{~eV} \mathrm{~nm}, \mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s} mass of the hydrogenatom =1.6×10−27 kg=1.6 \times 10^{-27} \mathrm{~kg} ]

Step-by-Step Verified Solution

ΔEamp;=13.6eV(14−19)amp;=6836eV=1.89eV\begin{aligned}\Delta E & =13.6 \mathrm{eV}\left(\frac{1}{4}-\frac{1}{9}\right) \\& =\frac{68}{36} \mathrm{eV}=1.89 \mathrm{eV}\end{aligned}

Due to recoil of hydrogen atom, the energy of emitted photon will decrease by very small amount.

So for approximate calculations,

amp;% charge =ΔEatom ΔE×100amp;=(ΔEC)22mΔE×100amp;=ΔEC2×2m×100amp;=1.89×1.6×10−19×100(3×108)2×2×1.6×10−27amp;=1.05×10−7%amp;∴n=7\begin{aligned}& \% \text { charge }= \frac{\Delta E_{\text {atom }}}{\Delta E} \times 100 \\&=\frac{\frac{\left(\frac{\Delta E}{C}\right)^2}{2 m}}{\Delta E} \times 100 \\&=\frac{\Delta E}{C^2 \times 2 m} \times 100 \\&=\frac{1.89 \times 1.6 \times 10^{-19} \times 100}{\left(3 \times 10^8\right)^2 \times 2 \times 1.6 \times 10^{-27}} \\&=1.05 \times 10^{-7} \% \\& \therefore n=7\end{aligned}

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