Question

An $\alpha$-particle moves with velocity $\frac{1}{30}$ times the velocity of light. If the uncertainty in position is $\frac{3.313}{\pi }$ pm, then the minimum uncertainty in kinetic energy is $y\times {10}^{-16}$ J. Then y is

Answer 1

$dKE=mvdv=mv\left(\frac{h}{4\pi \times m\times \Delta x}\right)$

$=m\times \frac{3\times {10}^8}{30}\times \frac{6.626\times {10}^{-34}\times \pi }{4\times \pi \times 3.313}$
$=5\times {10}^{-16}$