Two lines, A and B in the plot given below show the variation of de-Broglie wavelength, A versus $\frac{1}{\sqrt{V}},$ where V is the accelerating potential difference, for two particles carrying the same charge.
Which one of two represents a particle of smaller mass?

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Solution

D e - b r o g l i e w a v e l e n g t h i s g i v e n b y, λ = \frac{h}{p} λ = \frac{h}{mv} λ = \frac{1.227}{\sqrt{V}}

\Rightarrow

slope = \frac{λ}{1 / \sqrt{V}} = \frac{h / \sqrt{V}}{1 / \sqrt{V}}

slope = \frac{h}{mv} \times \frac{\sqrt{V}}{1.22}

S in c e, Slope of B > slope of A Slope \propto \frac{1}{m}

\frac{1}{\sqrt{m_{B}}} > \frac{1}{\sqrt{m_{A}}} \sqrt{m_{B}} < \sqrt{m_{A}}

∴

m_{B} < m_{A}

Therefore, line B represents a particle of smaller mass.

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