Question

An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom of mass $m$ acquired as a result of photon emission will be:
( $R$ is Rydberg constant and $h$ is Planck’s constant)
(A) $\dfrac{{25m}}{{24hR}}$
(B) $\dfrac{{24m}}{{25hR}}$
(C) $\dfrac{{24hR}}{{25m}}$
(D) $\dfrac{{25hR}}{{24m}}$

Answer 1

Hint: - Use the Rydberg formula to determine the wavelength of the emitted photon. At that moment use the formula for the energy of the photon to define the energy of the emitted photon. Use the relation between momentum and energy of the photon to find the velocity of the emitted photon.
Formula used: The Rydberg formula is,
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$ ...............$\left( 1 \right)$
Here, $\lambda $ is the wavelength of the photon emitted by an electron jumping from level ${n_2}$ to level ${n_1}$ and $R$is the Rydberg constant.
The energy $E$ of a photon is specified by
$E = \dfrac{{hc}}{\lambda }$ ............. $\left( 2 \right)$
Here, $h$ is Planck’s constant, $c$ is the speed of light, and $\lambda $ is the wavelength of the photon.
The energy $E$ of a photon in relations to momentum $P$ is
$P = \dfrac{E}{c}$ ............... $\left( 3 \right)$
Here, $c$ is the speed of light.
The momentum $P$ of an object given by
$P = mv$ ...............$\left( 4 \right)$
Here, $m$ is the mass of an object and $v$ is the velocity of the object.

Complete step-by-step answer:
An electron of a motionless hydrogen atom drives from the fifth energy level to the ground level.
We need to determine the wavelength $\lambda $ of the photon when the electron jumps from the fifth energy level to the ground level.
The ground level of the hydrogen atom is indicated by $1$ .
Substitute $1$ for ${n_1}$ and $5$ for ${n_2}$ is indicated by $1$.
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{5^2}}}} \right)$
$ \Rightarrow \dfrac{1}{\lambda } = R\left( {1 - \dfrac{1}{{25}}} \right)$
On further simplifying the above equation we get,
$\dfrac{1}{\lambda } = R\left( {\dfrac{{25 - 1}}{{25}}} \right)$
$ \Rightarrow \lambda = \dfrac{{25}}{{24R}}$
Determine the energy of the emitted photon.
Substitute $\dfrac{{25}}{{24R}}$ for $\lambda $ in the equation $\left( 2 \right)$ .
$E = \dfrac{{24Rhc}}{{25}}$
Substitute $\dfrac{E}{c}$ for $P$ in the equation $\left( 4 \right)$ .
$\dfrac{E}{c} = mv$
Substitute $\dfrac{{24Rhc}}{{25}}$ for $E$ in the above equation we get,
$\dfrac{{\dfrac{{24Rhc}}{{25}}}}{c} = mv$
$ \Rightarrow \dfrac{{24Rh}}{{25}} = mv$
Rearrange the above equation for the velocity $v$ of the emitted photon.
$v = \dfrac{{24hR}}{{25m}}$
As a result, the velocity of the emitted photon will be $\dfrac{{24hR}}{{25m}}$ .

Hence, the correct option is (C) $\dfrac{{24hR}}{{25m}}$ .

Note: One can also determine the velocity of the emitted photon using the law of conservation of linear momentum after determining the wavelength of the emitted photon. Ejection of electrons from a metal surface or the occurrence of a photoelectric effect happens only when the threshold frequency and threshold wavelength are met.