Question
A proton and an alpha particle are accelerated through the same potential. Which one of the two has, (i) greater value of de-Broglie wavelength associated with it and (ii) less kinetic energy. Give reasons to justify your answer.
✓
Answer
- Proton
$\lambda = \frac{\text{h}}{\sqrt{2\text{mqV}}}$
As mass of proton <mass of $\alpha$ particle and
$q_ ∝=2q_p$$\Rightarrow\lambda_{p} > \lambda_{∝}$ for the same accelerating potential.
- Alpha particle
$\text{K.E.} =\text{ qV}$
We have $q_p<q_{\alpha}$
$\therefore$ (For same accelerating potential)Kinetic energy of proton < KE of $\alpha$ particle.
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