Question

If the uncertainty in the position of an electron is ${10^{ - 10}}m$, then the value of uncertainty in its momentum will be ______________ $kgm{s^{ - 1}}$. $\left( {h = 6.62 \times {{10}^{ - 34}}J - s} \right)$.
(A) $0.53 \times {10^{ - 24}}$
(B) $1.35 \times {10^{ - 24}}$
(C) $1.06 \times {10^{ - 24}}$
(D) $1.08 \times {10^{ - 24}}$

Answer 1

Hint: To solve this question we need to use the Heisenberg’s uncertainty principle. The value of the uncertainty in the position of the electron is given in the question. Substituting it in the Heisenberg’s relation, we can get the required value of the uncertainty in the momentum of the electron.
Formula used: The formula used to solve this question is given by
\[\Delta x\Delta p = \dfrac{h}{{4\pi }}\], here $\Delta x$ is the uncertainty in the position, $\Delta p$ is the uncertainty in the momentum, and $h$ is the Planck’s constant.

Complete step-by-step answer:
According to the Heisenberg’s uncertainty principle, it is impossible to measure the position and the momentum of a particle simultaneously with high precision.
We know from the Heisenberg’s uncertainty principle that the uncertainty in the position and the momentum of a particle are related by the equation
\[\Delta x\Delta p = \dfrac{h}{{4\pi }}\] …………………….(1)
According to the question, the uncertainty in the position of the electron is given as
$\Delta x = {10^{ - 10}}m$ …………..(2)
Also the value of the Planck’s constant is given as
$h = 6.62 \times {10^{ - 34}}J - s$ ……………...(3)
Substituting (2) and (3) in (1) we get
\[{10^{ - 10}}\Delta p = \dfrac{{6.62 \times {{10}^{ - 34}}}}{{4\pi }}\]
Multiplying by ${10^{ - 10}}$ on both the sides we get
\[\Delta p = \dfrac{{6.62 \times {{10}^{ - 24}}}}{{4\pi }}kgm{s^{ - 1}}\]
On solving we get
$\Delta p = 0.53 \times {10^{ - 24}}kgm{s^{ - 1}}$
Thus, the value of the uncertainty in the momentum of the electron is equal to \[0.53 \times {10^{ - 24}}kgm{s^{ - 1}}\].

Hence, the correct answer is option A.

Note: The Heisenberg’s uncertainty principle can be used to prove the non existence of an electron in the nucleus. So we can prove that an electron can always revolve around the nucleus. Also, it is used in determining the Bohr’s radius, and the binding energy of the electron in an orbit.