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A particle is dropped from a height $\mathrm{H}$. The de-Broglie wavelength of the particle as a function of height is proportional to
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Verified Answer
The correct answer is:
$\mathrm{H}^{-1 / 2}$
$\mathrm{H}^{-1 / 2}$
Velocity of a body freely falling from a height $\mathrm{H}$ is
$$
v=\sqrt{2 \mathrm{gH}}
$$
So, $\quad \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{m} \sqrt{2 \mathrm{gH}}} \Rightarrow=\frac{\mathrm{h}}{\mathrm{m} \sqrt{2 \mathrm{~g}} \sqrt{\mathrm{H}}}$
(h, $\mathrm{m}$ and $g$ are constant)
Here, $\frac{\mathrm{h}}{\mathrm{m} \sqrt{2 \mathrm{~g}}}$ is also constant
So, $\mathrm{h} \propto \frac{1}{\sqrt{\mathrm{H}}} \Rightarrow$ or $\lambda \propto \mathrm{H}^{-1 / 2}$
$$
v=\sqrt{2 \mathrm{gH}}
$$
So, $\quad \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{m} \sqrt{2 \mathrm{gH}}} \Rightarrow=\frac{\mathrm{h}}{\mathrm{m} \sqrt{2 \mathrm{~g}} \sqrt{\mathrm{H}}}$
(h, $\mathrm{m}$ and $g$ are constant)
Here, $\frac{\mathrm{h}}{\mathrm{m} \sqrt{2 \mathrm{~g}}}$ is also constant
So, $\mathrm{h} \propto \frac{1}{\sqrt{\mathrm{H}}} \Rightarrow$ or $\lambda \propto \mathrm{H}^{-1 / 2}$
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