An electron of a stationary Hydrogen atom tra | vedclass

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Question

(A) 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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$\frac{24 Rh}{25 m}$

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(A) The wavelength $\lambda$ of the emitted photon is given by the Rydberg formula:$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$Given $n_1 = 1$ and $n_2 = 5$:$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{5^2} ight) = R \left( 1 - \frac{1}{25} ight) = \frac{24}{25} R$ ... $(i)$The momentum of the emitted photon is $p = \frac{h}{\lambda}$.Substituting from $(i)$:$p = h \left( \frac{24}{25} R ight)$By the law of conservation of linear momentum,the momentum of the atom must be equal and opposite to the momentum of the photon (since the atom was initially stationary):$p_{	ext{atom}} = p_{	ext{photon}}$$mv = \frac{24 Rh}{25}$$v = \frac{24 Rh}{25 m}$

$1$. 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 transitions from the fifth energy level to the ground level. The velocity that the atom acquires as a result of photon emission is ( $m=$ mass of the atom,$R=$ Rydberg's constant,$h=$ Planck's constant).

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