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An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be (m is the mass of the electron, R, Rydberg constant and h, Planck's constant)
A
$\frac{24m}{25hR}$
B
$\frac{24hR}{25m}$
C
$\frac{25hR}{24m}$
D
$\frac{25m}{24hR}$
Detailed Solution
Energy of the emitted photon: $E = hcR\left(\frac{1}{1^2} - \frac{1}{5^2}\right) = \frac{24}{25}hcR$
Momentum of the photon: $p = \frac{E}{c} = \frac{24hR}{25}$
By conservation of momentum the atom recoils with an equal momentum.
Velocity $= \frac{p}{m} = \frac{24hR}{25m}$
Momentum of the photon: $p = \frac{E}{c} = \frac{24hR}{25}$
By conservation of momentum the atom recoils with an equal momentum.
Velocity $= \frac{p}{m} = \frac{24hR}{25m}$