Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter $10^{-15} \mathrm{~m}$ is ________ $\times 10^9 \mathrm{~ms}^{-1}$ (nearest integer)

[Given : mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$, Plank's constant $(h)=6.626 \times 10^{-34} \mathrm{Js}$] (Value of $\pi=3.14$)

<p>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:</p> <p>$ \Delta x \cdot m \Delta v \geq \frac{h}{4 \pi} $</p> <p>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.</p> <p>Given:</p> <ul> <li>$ \Delta x = 10^{-15} \ \text{m} $</li><br> <li>$ m = 9.1 \times 10^{-31} \ \text{kg} $</li><br> <li>$ h = 6.626 \times 10^{-34} \ \text{Js} $</li><br> <li>$ \pi = 3.14 $</li> </ul> <p>We need to find $\Delta v$. Rearrange the uncertainty principle to solve for $\Delta v$:</p> <p>$ \Delta v \approx \frac{h}{4 \pi m \Delta x} $</p> <p>Plug in the numbers:</p> <p>$ \Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-15}} $</p> <p>First, calculate the denominator:</p> <p>$ 4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-15} = 114.392 \times 10^{-46} $</p> <p>Now, perform the division:</p> <p>$ \Delta v = \frac{6.626 \times 10^{-34}}{114.392 \times 10^{-46}} = \frac{6.626}{114.392} \times 10^{12} $</p> <p>Perform the division:</p> <p>$ \Delta v \approx 0.0579 \times 10^{12} \ \text{m/s} = 57.97 \times 10^9 \ \text{m/s} $</p> <p>So, the uncertainty in the velocity of the electron is:</p> <p>$ 58 \times 10^{9} \ \text{m/s} \quad \text{(nearest integer)} $</p>