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 m1, m2 (m1 > m2). 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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