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(A) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$.Since the kinetic energy $K = \frac{p^2}{2m} = qV$,the momentum is $p = \s

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$m_4$

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(A) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$.Since the kinetic energy $K = \frac{p^2}{2m} = qV$,the momentum is $p = \sqrt{2mqV}$.Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mqV}} = \left( \frac{h}{\sqrt{2mq}} ight) \left( \frac{1}{\sqrt{V}} ight)$.Comparing this with the equation of a straight line $y = mx$,where $y = \lambda$ and $x = \frac{1}{\sqrt{V}}$,the slope is given by $	ext{slope} = \frac{h}{\sqrt{2mq}}$.Since $h$ and $q$ are constants,the slope is inversely proportional to the square root of the mass,i.e.,$	ext{slope} \propto \frac{1}{\sqrt{m}}$.Therefore,a larger slope corresponds to a smaller mass.From the graph,the line corresponding to $m_4$ has the largest slope.Thus,$m_4$ represents the particle with the smallest mass.

The graph shows the variation of de-Broglie wavelength $\lambda$ versus $\frac{1}{\sqrt{V}}$,where $V$ is the accelerating potential for four particles carrying the same charge but having masses $m_1, m_2, m_3, m_4$. Which one represents the particle with the smallest mass?

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