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Question

(C) The wavelength $\lambda$ of the emitted photon is given by the Rydberg formula:$\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \ri

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$\frac{24hR}{25m}$

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(C) The wavelength $\lambda$ of the emitted photon is given by the Rydberg formula:$\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] = R \left[ \frac{1}{1^2} - \frac{1}{5^2} \right] = R \left[ 1 - \frac{1}{25} \right] = \frac{24}{25} R$.According to the law of conservation of momentum,the momentum of the atom $p_{atom}$ must be equal in magnitude to the momentum of the emitted photon $p_{photon}$ because the atom was initially stationary:$p_{atom} = p_{photon} = \frac{h}{\lambda}$.Since $p_{atom} = mv$,we have:$mv = \frac{h}{\lambda}$.Substituting the value of $\frac{1}{\lambda}$:$v = \frac{h}{m} \cdot \frac{1}{\lambda} = \frac{h}{m} \cdot \left( \frac{24}{25} R \right) = \frac{24hR}{25m}$.

$A$ stationary hydrogen atom has an electron that transitions from the fifth energy level to the ground level. The velocity that the atom acquires as a result of photon emission will be: ($m$ is the mass of the atom,$R$ is Rydberg constant,and $h$ is Planck's constant).

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