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MHT CET 2023 (12 May, Shift 2) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from MHT CET 2023 (12 May, Shift 2)

Q72MHT CET 2023 · 12 MAY · SHIFT 2Medium

Matter Waves: de Broglie Wavelength

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 same charge but of masses $\mathrm{m_1, m_2, m_3, m_4}$ . Which on represents a particle of largest mass?

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Answer: (a)

$\lambda = \dfrac{h}{\sqrt{2mq}} \cdot \dfrac{1}{\sqrt{V}}$ : a straight line through the origin with slope $\dfrac{h}{\sqrt{2mq}}$ .
Same $q$ , so slope $\propto \dfrac{1}{\sqrt{m}}$ : the largest mass has the smallest slope.
The line for $m_1$ is the least steep.
So the answer is (a) $m_1$ .

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