Question
A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has,
  1. greater value of de-Broglie wavelength associated with it, and
  2. less momentum?
Give reasons to justify your answer.

Answer

  1. de Broglie wavelength is given by
$\lambda = \frac{\text{h}}{\sqrt{2\text{mqV}}}$

As mass of proton < mass of deuteron and $q_p=q_d$ and v is same

$\Rightarrow\lambda_{p} > \lambda_{d}$ for same accelerating potential.
  1. Momentum $ = \frac{\text{h}}{\lambda}$
$\because\lambda_{p}> \lambda_{d}$

$\therefore$ momentum o proton will be less, than that of deuteron.

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