Question: Q. 5. Figure shows a plot of $\frac{1}{\sqrt{V}}$, where $V$ is the accelerating potential. The de-Broglie wavelength ’ $\lambda$ ’ in the case of two particles having same charge ’ $q$ ’ but different masses $m_{1}$ and $m_{2}$. Which line $(A$ or $B$ ) represents a particle of larger mass?
[OD Comptt. I, II, III 2013]
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Solution:
Ans.
$$ \lambda=\frac{h}{\sqrt{2 m e V}} $$
Line-Slope $\left(\frac{\frac{1}{\sqrt{V}}}{\lambda}\right) \propto \sqrt{m}$
Since, $\quad\left(\frac{\frac{1}{\sqrt{V}}}{\lambda}\right){A}>\left(\frac{\frac{1}{\sqrt{V}}}{\lambda}\right){B}$
Hence, line A will represent larger mass.
$$ \text { i.e., } \quad m_{1}>m_{2} \text { or } m_{A}>m_{B} \text {. } $$
