Question

An electron is allowed to move freely in a closed cubic box of length of side 10 $mm$. The uncertainity in its velocity will be:

• A. $3.35\times {10}^{-3}{ms}^{-1}$
• B. $5.8\times {10}^{-4}{ms}^{-1}$
• C. $4\times {10}^{-5}{ms}^{-1}$
• D. $4\times {10}^{-6}{ms}^{-1}$

Answer 1

The correct option is A $3.35\times {10}^{-3}{ms}^{-1}$

We know that from Heisenberg's uncertainity principle,
$\Delta x.\Delta p\ge \frac{h}{4\pi }$
where $\Delta x:\text{uncertainity in position}$
$\Delta p:\text{uncertainity in momentum}$
$h:\text{planck's constant}$
Side of cube $=10mm=10\times {10}^{-3}m$
Maximum uncertainity in position = Length of body diagonal of cube. Because it is the maximum possible length in the cube.
$\therefore \Delta x=10\sqrt{3}mm=10\sqrt{3}\times {10}^{-3}m$
Here, $p=m\text{v}$
$\therefore \Delta p=\Delta \left(m\text{v}\right)$
$\Rightarrow$ $\Delta p=m\Delta \text{v}$
$\therefore \Delta \text{v}\ge \frac{h}{4\pi m\Delta x}$
$\ge \frac{6.626\times {10}^{-34}}{4\times 3.14\times (9.1\times {10}^{-31})\times (10\sqrt{3}\times {10}^{-3})}$
$\therefore \Delta \text{v}\ge 3.34\times {10}^{-3}m/s$