Question

A stationary hydrogen atom of mass $M$ emits a photon corresponding to the first line of the Lyman series. If $R$ is the Rydberg's constant, the velocity that the atom acquires will be,
$\begin{aligned} & A.\dfrac{3}{4}\dfrac{Rh}{M} \\\ & B.\dfrac{Rh}{4M} \\\ & C.\dfrac{Rh}{2M} \\\ & D.\dfrac{Rh}{M} \\\ \end{aligned}$

Answer 1

The first line of the Lyman series is occurring when a transition from second shell to the first shell is happening. The momentum of the photon emitted should be calculated first. It is given by the ratio of the Planck’s constant to the wavelength which is equal to the product of the Rydberg constant and the difference of reciprocal of the square of the number of shells from one. Then apply the momentum conservation theorem and find out the velocity by rearranging the equation. These details will help you to solve this question.

Complete answer:
It is mentioned in the question that the transition is taking place from the second shell to the first one. That is,
$n=2\to n=1$
The momentum with which the photon is emitted can be calculated by the equation given as,
${{P}_{photon}}=\dfrac{h}{\lambda }=Rh\left( 1-\dfrac{1}{{{n}^{2}}} \right)$
Where $h$be the Planck’s constant, $\lambda$ be the wavelength of the light, $R$ be Rydberg constant and $n$ be the number of shells.
Substituting the value of $n$ will give,
${{P}_{photon}}=Rh\left( 1-\dfrac{1}{{{2}^{2}}} \right)=\dfrac{3Rh}{4}$
According to the momentum conservation, we can write that,
${{P}_{photon}}={{P}_{atom}}$
That is,
$MV=\dfrac{3Rh}{4}$
Where $M$ be the mass of the atom and $V$ be the velocity of the atom.
Rearranging the equation will give,
$V=\dfrac{3Rh}{4M}$
Therefore, the correct answer is obtained as option A.

Note: The Lyman series is defined as a series of hydrogen spectrum which will result in emission of light with a certain wavelength when the electron goes from the second shell to the first shell of an atom which is the lowest energy level of the electron. The wavelengths in the series of Lyman are in the range of ultraviolet.