Q42 / 104 · NTA Abhyas (JEE Main) 2020 · Physics · Atomic Physics

An electron of a stationary hydrogen atom (mass = m) jumps from the fifth energy level to the ground level. As a result, the atom emits a photon and recoils with some velocity. The velocity that the atom acquires as a result of photon emission is [h = Planck's constant, R = Rydberg's constant]

  • Correct answer
    A
    24hR25m
  • B
    25hR24m
  • C
    24m25hR
  • D
    25m24hR

Solution

Here, E5−E1=hcλ
−(Rhc52−Rhc)=hcλ
(where R=Rydberg's constant, h=Planck's constant and c=speed of light in vacuum)

⇒2425R=1λ

But  P=hλ
⇒v=hmλ=24Rh25m

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

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