Question: Q. 4. Plot a graph showing variation of de-Broglie wavelength versus $\frac{1}{\sqrt{V}}$, where, $V$ is accelerating potential for two particles $A$ and $B$ carrying same charge but of masses $m_{1}, m_{2}\left(m_{1}>m_{2}\right)$. Which one of the two represents a particle of smaller mass and why?
U] [Delhi I, II, III 2016]
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Solution:
Ans. As
$$ \lambda=\frac{h}{\sqrt{2 m q V}} $$
As the charge of two particles is same, therefore
$$ \frac{\lambda}{\left(\frac{1}{\sqrt{\mathrm{V}}}\right)} \propto \frac{1}{\sqrt{m}} \text { i.e., slope } \propto \frac{1}{\sqrt{m}} $$
2
Hence, particle with lower mass $\left(m_{2}\right)$ will have greater slope.
[CBSE Marking Scheme 2016]
Commonly Made Error
- Most of the students couldn’t relate the slope values.
- Some students even couldn’t draw the graph.
