A proton and a deuteron are accelerated through the same accelera… - Vidyadip

Question

A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has,

greater value of de$-$Broglie wavelength associated with it, and
less momentum?

Give reasons to justify your answer.

✓

Answer

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.

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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