An electron of a stationary hydrogen atom pas | vedclass

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Question

(A) According to the Rydberg formula:$\frac{1}{\lambda} = R \left[ \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right]$Here,$n_{f} = 1$ and $n_{i} = 5$.

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

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(A) According to the Rydberg formula:$\frac{1}{\lambda} = R \left[ \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} ight]$Here,$n_{f} = 1$ and $n_{i} = 5$.Substituting the values:$\frac{1}{\lambda} = R \left[ \frac{1}{1^{2}} - \frac{1}{5^{2}} ight] = R \left[ 1 - \frac{1}{25} ight] = \frac{24}{25} R$According to the law of conservation of linear momentum,the momentum of the emitted photon must be equal to the momentum acquired by the atom:$p_{	ext{photon}} = p_{	ext{atom}}$$\frac{h}{\lambda} = mv$Solving for velocity $v$:$v = \frac{h}{m\lambda} = \frac{h}{m} \left( \frac{24R}{25} ight) = \frac{24hR}{25m}$

$1$. Burning candle	$i$. Line spectrum
$2$. Sodium vapour	$ii$. Continuous spectrum
$3$. Bunsen flame	$iii$. Band spectrum
$4$. Dark lines in solar spectrum	$iv$. Absorption spectrum

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

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