Question

Plot a graph showing variation of de-broglie wavelength $\lambda$ versus $\frac{1}{\sqrt{V}}$ where v is acclerating 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 1

The de-Broglie wavelength is given by

$\lambda =\frac{h}{\sqrt{2mqV}}$

The graph of $\lambda$ vs $\frac{1}{\sqrt{V}}$ is a straight line with slope $\frac{h}{\sqrt{2mq}}$

Since charge $q$ of A and B is same , slope is $\frac{1}{\sqrt{m}}$

Since, $m1>m2$

$\text{Slope of B}<\text{Slope of A}$

More is the slope, less is the mass, since slope of A is larger .So plot A represents smaller mass $m2$