Question

A particle is dropped from a height $H$. The de Broglie wavelength of the particle as a function of height is proportional to

• A. $H^{\frac{-1}{2}}$
• B. $H^0$
• C. $H$
• D. $H^{\frac{1}{2}}$

Answer 1

The correct option is A $H^{\frac{-1}{2}}$

The wave associated with a moving particle is known as de-Broglie wave.
The wavelength of de-Broglie wave is given by, $\lambda =\frac{h}{p}$

$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mE}}$

Here,$h=\text{Plank's constant}$
$v=\text{speed of the particle}$
$E=\text{energy of the particle}$
For a body falling freely under gravity from a height $H$,
Let the final velocity of the particle be $v$.
Using Newton's third equation of motion,
$v^2-u^2=2gH$
$v=\sqrt{2gH}$ $(u=0$)
So, de-Broglie wavelength,
$\lambda =\frac{h}{m\sqrt{2gH}}$
$\lambda \propto H^{\frac{-1}{2}}$
Final answer: $\left(d\right)$