1. Use Heisenberg's uncertainty principle

For position and momentum uncertainty:

$\Delta x\,\Delta p \ge \frac{h}{4\pi}$

Since

$\Delta p = m\,\Delta v$

we get

$\Delta x\,m\,\Delta v \ge \frac{h}{4\pi}$

So,

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

2. Substitute the given values

• $h = 6.626 \times 10^{-34}\,\text{Js}$
• $m = 9.1 \times 10^{-31}\,\text{kg}$
• Diameter of nucleus $= 10^{-15}\,\text{m}$

If electron is confined within the nucleus, take

$\Delta x = 10^{-15}\,\text{m}$

Thus,

$\Delta v = \frac{6.626 \times 10^{-34}}{4\times 3.14 \times 9.1 \times 10^{-31} \times 10^{-15}}$

3. Calculate the denominator

$4\pi = 4 \times 3.14 = 12.56$

$12.56 \times 9.1 = 114.296$

So denominator is

$114.296 \times 10^{-46} = 1.14296 \times 10^{-44}$

4. Now calculate $\Delta v$

$\Delta v = \frac{6.626 \times 10^{-34}}{1.14296 \times 10^{-44}}$

$\Delta v = \left(\frac{6.626}{1.14296}\right) \times 10^{10}$

$\Delta v \approx 5.80 \times 10^{10}\,\text{m s}^{-1}$

This can be written as

$\Delta v \approx 58 \times 10^9\,\text{m s}^{-1}$

5. Nearest integer

The blank is:

$\boxed{58}$

6. Comparison with stored answer

Stored correct answer = 58

Our derived answer also = 58, so they agree.