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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)
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( $m$ is the mass of the electron, $R$ Rydberg constant and $h$ Planck's constant)
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1628 Upvotes
Verified Answer
The correct answer is:
$\frac{24 h R}{25 m}$
Here, $E_5-E_1=\frac{h c}{\lambda}$
$\begin{gathered}
\frac{R h c}{25}-R h c=\frac{h c}{\lambda} \\
\frac{24}{25} R=\frac{1}{\lambda}
\end{gathered}$
But
$\begin{aligned}
P & =\frac{h}{\lambda} \\
v & =\frac{h}{m \lambda} \\
& =\frac{24}{25} \frac{R h}{m}
\end{aligned}$
$\begin{gathered}
\frac{R h c}{25}-R h c=\frac{h c}{\lambda} \\
\frac{24}{25} R=\frac{1}{\lambda}
\end{gathered}$
But
$\begin{aligned}
P & =\frac{h}{\lambda} \\
v & =\frac{h}{m \lambda} \\
& =\frac{24}{25} \frac{R h}{m}
\end{aligned}$
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