When a hydrogen atom going from n=2 to n=1 emits a photon, its recoil speed is \fracx5 ~m / s . Where x= \underline\hspace2cm . (Use, mass of hydrogen atom =1.6 \times 10^-27 ~kg )

Atoms and Nuclei PYQ — JEE Main 2024 · 29 Jan · Shift 1 · Q84 with Solution

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Correct answer: 17

For hydrogen, the energy levels are $E_n=-\frac{13.6}{n^2}\text{ eV}$

For the transition $n=2 \to n=1$ , $\Delta E = 13.6\left(1-\frac{1}{4}\right)=13.6\cdot \frac{3}{4}=10.2\text{ eV}$

So the emitted photon has energy $E_\gamma=10.2\text{ eV}$

Converting to joule: $E_\gamma=10.2\times 1.6\times 10^{-19} = 1.632\times 10^{-18}\text{ J}$

Photon momentum is $p_\gamma=\frac{E_\gamma}{c}$

Taking $c=3\times 10^8\,\text{m/s}$ ,

=5.44\times 10^{-27}\,\text{kg m/s}$$ 3. **Recoil momentum of hydrogen atom** By conservation of momentum, the atom recoils with equal momentum: $$p_{\text{atom}}=p_\gamma$$ If recoil speed is $v$, then $$Mv=p_\gamma$$ Given mass of hydrogen atom, $$M=1.6\times 10^{-27}\,\text{kg}$$ So, $$v=\frac{5.44\times 10^{-27}}{1.6\times 10^{-27}}=3.4\,\text{m/s}$$ 4. **Match with given form** Given recoil speed is $$\frac{x}{5}\,\text{m/s}$$ Thus, $$\frac{x}{5}=3.4$$ $$x=17$$ Hence, the required integer is $$\boxed{17}$$