Q141 / 200 · NEET 2012 (Screening) · Physics

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)

  • Correct answer
    A
    $\frac{24 h R}{25 m}$
  • B
    $\frac{25 h R}{24 m}$
  • C
    $\frac{25 m}{24 h R}$
  • D
    $\frac{24 m}{25 h R}$

Solution

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

Hence, the correct answer is option: 'A'.

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