Answer:
de- Broglie wavelength, \[\lambda =\frac{h}{\sqrt{2mqV}}=\frac{h}{\sqrt{2mq}}.\frac{1}{\sqrt{V}}\] \[\therefore \] Slope of \[\lambda \]versus \[1/\sqrt{V}\]graph \[=\frac{h}{\sqrt{2mq}}\] For the particle of same charge \[\text{q}\]. Slope \[\propto \frac{1}{\sqrt{m}}\] As the slope of line A is smaller than that of line B, so the line A represents the heavier particle.
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