Question

An $\alpha -$ particle is moving with a velocity $\frac{1}{30}th$ of the velocity of light. If the uncertainty in position is $\frac{3.31}{\pi }\text{pm}$, then the minimum uncertainty in the kinetic energy is $y\times {10}^{-16}\text{J}$. Calculate the value of $y$

Answer 1

$d\left(KE\right)=mvdv=mv\frac{h}{4\pi m\Delta x}$
$=\frac{3\times {10}^8}{3}\times \frac{6.62\times {10}^{-34}}{4\times \pi \times \frac{3.31}{\pi }\times {10}^{-12}}=5\times {10}^{-16}J$