Question

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

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

Answer 1

Answer: (B)

${5.8\times 10}^{-4}mse{c}^{-1}$

Given that

The electron is freely in a closed cubic box (CCB) of length $\left(a\right)=10cm$

Also given that

The electron is freely moving inside the box.

Therefore,

The maximum uncertainly in position will be equal to the length of the diagonal.

$\Delta x=\sqrt{3a}$

So,

$\Delta x=\sqrt{3}\times 10$

$\Delta x\left(m\Delta v\right)=\frac{h}{4\pi }$

$\Delta v=\frac{h}{4\pi \left(m\right)\Delta x}$

$\Rightarrow \frac{6.623\times {10}^{-34}}{10\sqrt{3}\times {10}^3\times 9.1\times {10}^{-31}\times 4\times 3.14}$

$\Rightarrow 5\times {10}^{-4}m{s}^{-1}$

Hence,

The maximum uncertainty in its velocity $5\times {10}^{-4}m{s}^{-1}$