MCQ
Free ${}^{238}U$ nuclei kept in a train emit alpha particles. When the train is stationary and a uranium nucleus decays, a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes $x$ in time $t$ after the decay. If a decay takes place when the train is moving at a uniform speed $v,$ the distance between the alpha particle and the recoiling nucleus at a time $t$ after the decay, as measured by the passenger will be:
  • A
    $x + vt$
  • B
    $x - vt$
  • $x$
  • D
    depends on the direction of the train.

Answer

Correct option: C.
$x$
When the train is stationary, the separation between the alpha particle and recoiling uranium nucleus is $x$ in time $t$ after the decay. Even if the decay is taking place in a moving train and the separation is measured by the passenger sitting in it, the separation between the alpha particle and nucleus will be $x.$ This is because the observer is also moving with the same speed with which the alpha particle and recoiling nucleus are moving, i.e. they all are in the same frame that is moving at a uniform speed.

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