Question: Q. 6. A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has :
(i) greater value of de-Broglie wavelength associated with it, and
(ii) less momentum ?
Give reasons to justify your answer.
U] [Delhi I, II, III 2014]
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Solution:
Ans. (i) de-Broglie wavelength is given by
$$ \lambda=\frac{h}{\sqrt{2 m q V}} $$
As mass of proton $<$ mass of deuteron and $q_{p}=q_{d}$ and $V$ is same.
$\therefore \lambda_{p}>\lambda_{d}$ for same accelerating potential.
$$ \begin{equation*} \text { Momentum }=\frac{h}{\lambda} \tag{ii} \end{equation*} $$
$$ \lambda_{p}>\lambda_{d} $$
$\therefore$ Momentum of proton will be less than that of deuteron.
[CBSEMarking Scheme 2014]
