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(B) The recoil momentum of the $H$ atom is equal to the momentum of the emitted photon.The momentum of a photon is given by $p = \frac{h}{\lambda}$.Us

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$6.8 	imes 10^{-27} 	ext{ N} \cdot 	ext{s}$

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(B) The recoil momentum of the $H$ atom is equal to the momentum of the emitted photon.The momentum of a photon is given by $p = \frac{h}{\lambda}$.Using the Rydberg formula for the wavenumber: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Substituting this into the momentum equation: $p = hR \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Given $n_1 = 1$ and $n_2 = 4$,we have:$p = hR \left( \frac{1}{1^2} - \frac{1}{4^2} ight) = hR \left( 1 - \frac{1}{16} ight) = \frac{15hR}{16}$.Using $h = 6.63 	imes 10^{-34} 	ext{ J} \cdot 	ext{s}$ and $R = 1.097 	imes 10^7 	ext{ m}^{-1}$:$p = \frac{15 	imes (6.63 	imes 10^{-34}) 	imes (1.097 	imes 10^7)}{16} \approx 6.8 	imes 10^{-27} 	ext{ kg} \cdot 	ext{m/s}$ (or $	ext{N} \cdot 	ext{s}$).

In a hydrogen atom,an electron makes a transition from the $n=4$ to the $n=1$ energy level. The recoil momentum of the $H$ atom will be:

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