QJEEM24Chdfb4: JEE Main 2024 Chemistry - Class 11 JEEM24Chdfb4 Solution: Based on Heisenberg's uncertainty principle, the uncertainty in …

To find the uncertainty in the velocity of the electron within an atomic nucleus of diameter $10^{-15} \ \text{m}$ , we can use Heisenberg's uncertainty principle. The principle is expressed as:

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

Here, $\Delta x$ is the uncertainty in position, $m$ is the mass of the electron, $\Delta v$ is the uncertainty in velocity, and $h$ is Planck's constant.

Given:

• $\Delta x = 10^{-15} \ \text{m}$


• $m = 9.1 \times 10^{-31} \ \text{kg}$


• $h = 6.626 \times 10^{-34} \ \text{Js}$


• $\pi = 3.14$

We need to find $\Delta v$ . Rearrange the uncertainty principle to solve for $\Delta v$ :

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

Plug in the numbers:

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

First, calculate the denominator:

$4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-15} = 114.392 \times 10^{-46}$

Now, perform the division:

$\Delta v = \frac{6.626 \times 10^{-34}}{114.392 \times 10^{-46}} = \frac{6.626}{114.392} \times 10^{12}$

Perform the division:

$\Delta v \approx 0.0579 \times 10^{12} \ \text{m/s} = 57.97 \times 10^9 \ \text{m/s}$

So, the uncertainty in the velocity of the electron is:

$58 \times 10^{9} \ \text{m/s} \quad \text{(nearest integer)}$