A particle is dropped from height H. The de-Broglie wavelength of the particle as a function of height is proportional to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $H^{-1/2}$
The De-Broglie Wavelength is -
$λ=\frac{h}{momentum (P)}$
and,
Momentum, $P=mv$
$=m\sqrt{2g(H-h)}$ $[v=\sqrt{2g(H-h)}]$
Hence,
$λ=\frac{h}{m\sqrt{2g(H-h)}}⇒λ∝\frac{1}{\sqrt{H}}$