JEE Main 2024 — Structure Of Atom Question with Solution

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:

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

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:

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