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(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Case $A$: For free fall, $v = \sqrt{2gh}$. Since $v$ is independent o

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Both move with the same speed

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(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Case $A$: For free fall, $v = \sqrt{2gh}$. Since $v$ is independent of mass, $\lambda = \frac{h}{m\sqrt{2gh}} \propto \frac{1}{m}$. Thus, the heavier particle has a smaller wavelength.Case $B$: With same kinetic energy $K$, $\lambda = \frac{h}{\sqrt{2mK}}$. Since $\lambda \propto \frac{1}{\sqrt{m}}$, the heavier particle has a smaller wavelength.Case $C$: With same momentum $p$, $\lambda = \frac{h}{p}$. Since $p$ is the same, $\lambda$ is the same for both.Case $D$: With same speed $v$, $\lambda = \frac{h}{mv}$. Since $\lambda \propto \frac{1}{m}$, the heavier particle has a smaller wavelength.Note: Options $A$, $B$, and $D$ all result in the heavier particle having a smaller de-Broglie wavelength. However, in standard physics problems of this type, the most direct relationship is often associated with constant speed or constant kinetic energy. Given the options, $D$ is the most fundamental case where $\lambda \propto 1/m$ is directly observed.

Consider two particles of different masses. In which of the following situations will the heavier of the two particles have a smaller de-Broglie wavelength?

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