Question

An electron of stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom of mass \[m\] acquired 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 Rydberg formula to determine the wavelength of the emitted photon. Then use the formula for energy of the photon to determine the energy of the emitted photon. Use the relation between momentum and energy of the photon to determine the velocity of the emitted photon.

Formula Used: The expression for the Rydberg formula is
\[\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)\] …… (1)

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 given by
\[E = \dfrac{{hc}}{\lambda }\] …… (2)

Here, \[h\] is Planck’s constant, \[c\] is the speed of light and \[\lambda \] is the wavelength of the photon.

The energy \[E\] of photon in terms of momentum \[P\] is
\[P = \dfrac{E}{c}\] …… (3)

Here, \[c\] is the speed of light.

The momentum \[P\] of an object is given by
\[P = mv\] …… (4)

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 stationary hydrogen atom passes from the fifth energy level to the ground level.

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 denoted by 1.

Substitute 1 for \[{n_1}\] and 5 for \[{n_2}\] in equation (1).
\[\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{5^2}}}} \right)\]
\[ \Rightarrow \dfrac{1}{\lambda } = R\left( {1 - \dfrac{1}{{25}}} \right)\]
\[ \Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{{25 - 1}}{{25}}} \right)\]
\[ \Rightarrow \lambda = \dfrac{{25}}{{24R}}\]

Hence, the wavelength of the emitted photon is \[\dfrac{{25}}{{24R}}\].

Determine the energy of the emitted photon.

Substitute \[\dfrac{{25}}{{24R}}\] for \[\lambda \] in equation (2).
\[E = \dfrac{{24Rhc}}{{25}}\]

Substitute \[\dfrac{E}{c}\] for \[P\] in equation (4).
\[\dfrac{E}{c} = mv\]

Substitute \[\dfrac{{24Rhc}}{{25}}\] for \[E\] in the above equation.
\[\dfrac{{\dfrac{{24Rhc}}{{25}}}}{c} = mv\]
\[ \Rightarrow \dfrac{{24Rh}}{{25}} = mv\]

Rearrange the above equation for the velocity \[v\] of the emitted photon.
\[v = \dfrac{{24hR}}{{25m}}\]

Therefore, the velocity of the emitted photon will be \[\dfrac{{24hR}}{{25m}}\].

Hence, the correct option is C.

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.