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(A) The energy of the emitted photon is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. Fo

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$\frac{15}{16} \frac{hR}{m}$

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(A) The energy of the emitted photon is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$. For a hydrogen atom $(Z=1)$ transitioning from $n_2 = 4$ to $n_1 = 1$: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{4^2} ight) = R \left( 1 - \frac{1}{16} ight) = \frac{15R}{16}$. The momentum of the emitted photon is $p = \frac{h}{\lambda}$. Substituting the value of $\frac{1}{\lambda}$: $p = h \left( \frac{15R}{16} ight) = \frac{15Rh}{16}$. By the law of conservation of momentum,the momentum of the hydrogen atom must be equal and opposite to the momentum of the photon: $p_{	ext{atom}} = p_{	ext{photon}} = mv$. Therefore,$mv = \frac{15Rh}{16}$,which gives $v = \frac{15}{16} \frac{Rh}{m}$.

An electron of a stationary hydrogen atom transits from the fourth energy level to the ground level. The velocity of the hydrogen atom acquired as a result of photon emission will be:
{ $R =$ Rydberg constant,$m =$ mass of hydrogen atom }

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An electron of a stationary hydrogen atom transits from the fourth energy level to the ground level. The velocity of the hydrogen atom acquired as a result of photon emission will be: { $R =$ Rydberg constant,$m =$ mass of hydrogen atom }