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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

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 is the Rydberg constant and h is the Planck's constant)

Option: 1

\frac{24hR}{25m}


Option: 2

\frac{25hR}{24m}


Option: 3

\frac{25m}{24hR}


Option: 4

\frac{24m}{25hR}


Answers (1)

best_answer

For hydrogen atom 

\begin{aligned} & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{\mathrm{f}}^2}-\frac{1}{\mathrm{n}_i^2}\right] \\ & \mathrm{n}_{\mathrm{f}}=1, \mathrm{n}_i=5 \\ & \frac{1}{\lambda}=\mathrm{R}\left[1-\frac{1}{(5)^2}\right]=\frac{24 R}{25} \end{aligned}

 

\therefore the momentum of emitted photon =  \frac{h}{\lambda }= momentum recoil of hydrogen atom=mv

  \begin{aligned} & \frac{\mathrm{h}}{\lambda}=m v \\ & v=\frac{h}{m \lambda}=\frac{h}{m} \times \frac{24 R}{25} \\ & v=\frac{24 h R}{25 m} \end{aligned}

Posted by

Sanket Gandhi

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