An electron of a stationary hydrogen atom makes a transition 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, $R$ is the Rydberg constant and $h$ is the Planck's constant)

\frac{24 h R}{25 m}

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\frac{25 h R}{24 m}

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\frac{25 m}{24 h R}

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\frac{24 m}{25 h R}

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Solution

The correct option is A $\frac{24 h R}{25 m}$
According to Rydberg formula
$\frac{1}{λ} = R [\frac{1}{{n_{f}}^{2}} - \frac{1}{{n_{i}}^{2}}]$
Here, $n_{f} = 1, n_{i} = 5$
$∴ \frac{1}{λ} = R [\frac{1}{1^{2}} - \frac{1}{5^{2}}] = R [\frac{1}{1} - \frac{1}{25}]$
$= \frac{24}{25} R$
According to conservation of linear momentum, we get
Momentum of photon = Momentum of atom
$\frac{h}{λ} = m v$
or $v = \frac{h}{m λ} = \frac{h}{m} (\frac{24 R}{25}) = \frac{24 h R}{25 m}$

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