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(C) The energy of the emitted photon during the transition from $n_2=5$ to $n_1=4$ is given by $\Delta E = h 
u = R h c \left( \frac{1}{n_1^2} - \fra

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$\frac{9 R h}{400 m}$

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(C) The energy of the emitted photon during the transition from $n_2=5$ to $n_1=4$ is given by $\Delta E = h 
u = R h c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Since the momentum of the photon is $p = \frac{E}{c} = \frac{h}{\lambda} = R h \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$,we substitute $n_1=4$ and $n_2=5$.$p = R h \left( \frac{1}{16} - \frac{1}{25} ight) = R h \left( \frac{25-16}{400} ight) = \frac{9 R h}{400}$.By the law of conservation of linear momentum,the recoil momentum of the atom must be equal in magnitude to the momentum of the emitted photon.Therefore,$m v = p$,where $v$ is the recoil speed.$v = \frac{p}{m} = \frac{9 R h}{400 m}$.

$A$ stationary hydrogen atom undergoes a transition from $n=5$ to $n=4$. The recoil speed of the atom is (where $R=$ Rydberg constant,$h=$ Planck's constant,and $m=$ mass of the hydrogen atom).

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