Question

The uncertainty in velocity of an electron present in the nucleus of diameter 10^-15 m hypothetically should be approximately ????

Answer 1

DATA USED
m of electron = 9.1x10^-31
Heisenberg uncertainty formula ∆x.m∆v=h/4π

Answer 2

Uncertainty velocity of an electron is $\bold{5.8 \times 10^{10} \ \mathrm{m} / \mathrm{s}}$

Explanation:

We cannot determine the position of electron in nucleus is known as uncertainty of an electron.

According to principle of Heisenberg uncertainty it is impossible for precisely determination of particles's velocity and position at the same time.

According to the uncertainty principle,

$\Delta x \cdot \Delta p=\frac{h}{4 \pi}$

$\Delta \mathrm{x}=\text { position of an electron }$

$\Delta \mathrm{p}=\text { momentum of an electron }$

$\mathrm{h}=\text { plank's constant }=6.626 \times 10^{-34}$

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

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

$\Delta \mathrm{v}=\frac{6.626 \times 10^{-34}}{4 \times \frac{22}{7} \times 10^{-15} \mathrm{m} \times 9.1 \times 10^{-31}}$

$\Delta \mathrm{v}=5.8 \times 10^{10} \ \mathrm{m} / \mathrm{s}$