Question

Calculate the uncertainty in the position ($\Delta x)$ of an electron if $\Delta vis0.1\%$. Take the velocity of electron $=2.2\times {10}^{6}m{s}^{-1}$ and mass of electron as $9.108\times {10}^{-31}kg$.

• A. $262.48\times {10}^{-10}m$
• B. $9.1\times {10}^{-9}m$
• C. $458.76\times {10}^{-12}m$
• D. $959.67\times {10}^{-10}m$

Answer 1

The correct option is A

$262.48\times {10}^{-10}m$

From Heisenberg Uncertainty Principle, we know

$\Delta p$ $\times$ $\Delta x$ = $\frac{h}{4\pi }$ $\Rightarrow$ $\Delta v$ $\times$ $\Delta x$ = $\frac{h}{4m\pi }$

Now,

Given , $\Delta v=0.1\%$ of the velocity of the electron

$=\frac{0.1}{100}$ $\times 2.2$ $\times$ ${10}^6$ $=2.2\times$ ${10}^{3}m{s}^{-1}$

$\Delta x\times m\Delta v=\frac{h}{4\pi }$

or $\Delta x=\frac{6.63\times {10}^{-34}Js}{4\times 3.14\times 9.108\times {10}^{-31}kg\times 2.2\times {10}^{3}m{s}^{-1}}$

$=0.02624765\times {10}^{-6}m$

$=262.4765\times {10}^{-10}m$

Since $\Delta x$ is much longer than the atomic diameter $(\approx$ ${10}^{-10}m)$, the uncertainty principle is applicable in this case.