Q25 / 90 · JEE Main 2024 (09 Apr Shift 2) · Chemistry
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 ________. \(10^9 \mathrm{~ms}^{-1}\) (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 \(\pi=3.14\))
Correct answer: 58
Solution
$\begin{aligned} & \mathrm{m} \Delta \mathrm{V} . \Delta \mathrm{x}=\frac{\mathrm{h}}{4 \pi} \\ & \begin{aligned} \Delta \mathrm{V} & =\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^{-15} \times 4 \times 3.14} \\ & =57.97 \times 10^{+9} \mathrm{~m} / \mathrm{sec}\end{aligned}\end{aligned}$
Hence, the correct answer is 58.