Graph shows the variation of de-Broglie wavelength $(\lambda)$ versus $\frac{1}{\sqrt{V}}$ where ' $V$ ' is…
Graph shows the variation of de-Broglie wavelength $(\lambda)$ versus $\frac{1}{\sqrt{V}}$ where ' $V$ ' is the accelerating potential for four particles A, B, C, $\mathrm{D}$ carrying same charge but of masses $\mathrm{m}_1, \mathrm{~m}_2$, $\mathrm{m}_3, \mathrm{~m}_4$. Which on represents a particle of largest mass?
$\mathrm{m}_1$
$\mathrm{m}_2$
$\mathrm{m}_3$
$\mathrm{m}_4$
Solution
de Broglie wayelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$
$\therefore \quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqv}}}$
$\therefore \quad \lambda \sqrt{\mathrm{v}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mq}}} \Rightarrow \frac{\lambda}{\left(\frac{1}{\sqrt{\mathrm{v}}}\right)}=\frac{1}{\sqrt{2 \mathrm{mq}}}$
$\therefore \quad$ Slope of the graph $=\frac{1}{\sqrt{2 \mathrm{mq}}}$
The slope will be maximum for minimum mass.
$\therefore \quad \mathrm{m}_4$ is minimum and $\mathrm{m}_1$ will be maximum.
