Question
Plot a graph showing variation of de$-$Broglie wavelength $\lambda$ versus $\frac{1}{\sqrt{V}} ,$ where $V$ is accelerating potential for two particles $A$ and $B$ carrying same charge but of masses $m_1, m_2 (m_1 > m_2).$ Which one of the two represents a particle of smaller mass and why?

Answer



As $\lambda = \frac{\hslash}{\sqrt{2\text{mqV}}}$
As the charge of two particles is same, therefore
$\frac{\lambda}{(\frac{1}{\sqrt{\text{V}}})} \alpha\frac{1}{\sqrt{\text{m}}}$  i.e.Slope $\alpha\frac{1}{\sqrt{\text{m}}}$
Hence, particle with lower mass ($\text{m}_{2}$) will have greater slope.

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