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(C) The hydrogen atom is initially at rest. By the law of conservation of momentum,the momentum of the emitted photon must be equal in magnitude to th

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$4$

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(C) The hydrogen atom is initially at rest. By the law of conservation of momentum,the momentum of the emitted photon must be equal in magnitude to the recoil momentum of the hydrogen atom.$p_{	ext{atom}} = p_{	ext{photon}}$$Mv = \frac{E}{c}$Where $E$ is the energy of the emitted photon,given by $E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) 	ext{eV}$.Substituting the values: $n_1 = 1$,$n_2 = 5$,$M = 1.6 	imes 10^{-27} \ kg$,and $c = 3 	imes 10^8 \ m/s$.$E = 13.6 	imes \left( 1 - \frac{1}{25} ight) 	imes 1.6 	imes 10^{-19} \ J = 13.6 	imes \frac{24}{25} 	imes 1.6 	imes 10^{-19} \ J$.$v = \frac{13.6 	imes 0.96 	imes 1.6 	imes 10^{-19}}{1.6 	imes 10^{-27} 	imes 3 	imes 10^8} = \frac{13.6 	imes 0.96 	imes 10^8}{3} = 4.352 \ m/s$.Thus,the recoil speed is approximately $4 \ m/s$.

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

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