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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$.Since $p = \sqrt{2mqV}$,we have $\lambda = \frac{h}{\sqrt{2mqV}}$.Rearranging the te

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$m_1$

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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$.Since $p = \sqrt{2mqV}$,we have $\lambda = \frac{h}{\sqrt{2mqV}}$.Rearranging the terms,we get $\lambda = \left( \frac{h}{\sqrt{2mq}} ight) \left( \frac{1}{\sqrt{V}} ight)$.Comparing this with the equation of a straight line $y = mx$,where $y = \lambda$ and $x = \frac{1}{\sqrt{V}}$,the slope of the graph 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: $	ext{Slope} \propto \frac{1}{\sqrt{m}}$.Therefore,a smaller slope corresponds to a larger mass.Looking at the graph,the line corresponding to $m_1$ has the smallest slope.Thus,$m_1$ represents the particle with the largest 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 $A, B, C, D$ carrying the same charge but having masses $m_1, m_2, m_3, m_4$. Which one represents a particle of the largest mass?

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