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JEE Main 2025 (29 Jan, Shift 1) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2025 (29 Jan, Shift 1)

Q29JEE Main 2025 · 29 JAN · SHIFT 1Easy

Matter Waves: de Broglie Wavelength

If $\lambda$ and $K$ are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be:

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Answer: (d)

$\lambda = \dfrac{h}{\sqrt{2mK}} \Rightarrow \dfrac{1}{K} = \dfrac{2m}{h^2}\lambda^2$ .
So $\dfrac{1}{K} \propto \lambda^2$ : a parabola through the origin, curving upward.
So the answer is (d).

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