Question

An $\alpha$ particle moving with velocity $\frac{1}{30}$ th times of velocity of light. If uncertainity in position is $\frac{3.31}{\pi }\text{pm}$, then minimum uncertainity in kinetic energy is:

• A. $5\times {10}^{-16}\text{J}$
• B. $2.5\times {10}^{-16}\text{J}$
• C. $5\times {10}^{-18}\text{J}$
• D. $2.5\times {10}^{-18}\text{J}$

Answer 1

The correct option is A $5\times {10}^{-16}\text{J}$

From Heisenburg's uncertainity principle, we know that
$\Delta p.\Delta x=m\Delta v.\Delta x=\frac{h}{4\pi }$
The minimum uncertainity in kinetic energy is
$d\left(\text{K.E.}\right)=d\left(\frac{1}{2}m{v}^2\right)=mv.dv=mv\frac{h}{4\pi m\Delta x}=\frac{3\times {10}^8}{30}\times \frac{6.62\times {10}^{-34}}{4\times \pi \times \frac{3.31}{\pi }\times {10}^{-12}}=5\times {10}^{-16}\text{J}$