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Physics Question on Atoms

A hydrogen atom emits a photon corresponding to an electron transition from n=5n=5 to n=1n=1. The recoil speed of hydrogen atom is almost (mass of proton ≈1.6×10−27 kg\approx 1.6 \times 10^{-27} \,kg )

A

10 ms−110\,ms^{-1}

B

2×10−2 ms−12 \times 10^{-2} \,ms ^{-1}

C

4 ms−14\, m s^{-1}

D

8×102 ms−18 \times 10^{2} \,m s^{-1}

Answer

4 ms−14\, m s^{-1}

Explanation

Solution

The hydrogen atom before the transition was at rest. Therefore, from conservation of momentum PH-atom=Pphoton=Eradiated c=13.6[1n12−1n22]eVcP_{\text{H-atom}}=P_{\text{photon}}=\frac{E_{\text {radiated }}}{c}=\frac{13.6\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] eV }{c} 1.6×10−27×v1.6 \times 10^{-27} \times v =13.6[112−152]×1.6×10−193×106=\frac{13.6\left[\frac{1}{1^{2}}-\frac{1}{5^{2}}\right] \times 1.6 \times 10^{-19}}{3 \times 10^{6}} ⇒v=4.352 m/s=≈4 m/s\Rightarrow v=4.352\, m / s =\approx 4\, m / s