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Question: If the uncertainty in the position of an electron is \({10^{ - 10}}m\), then the value of uncertaint...

If the uncertainty in the position of an electron is 10−10m{10^{ - 10}}m, then the value of uncertainty in its momentum will be ______________ kgms−1kgm{s^{ - 1}}. (h=6.62×10−34J−s)\left( {h = 6.62 \times {{10}^{ - 34}}J - s} \right).
(A) 0.53×10−240.53 \times {10^{ - 24}}
(B) 1.35×10−241.35 \times {10^{ - 24}}
(C) 1.06×10−241.06 \times {10^{ - 24}}
(D) 1.08×10−241.08 \times {10^{ - 24}}

Explanation

Solution

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
ΔxΔp=h4π\Delta x\Delta p = \dfrac{h}{{4\pi }}, here Δx\Delta x is the uncertainty in the position, Δp\Delta p is the uncertainty in the momentum, and hh 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
ΔxΔp=h4π\Delta x\Delta p = \dfrac{h}{{4\pi }} …………………….(1)
According to the question, the uncertainty in the position of the electron is given as
Δx=10−10m\Delta x = {10^{ - 10}}m …………..(2)
Also the value of the Planck’s constant is given as
h=6.62×10−34J−sh = 6.62 \times {10^{ - 34}}J - s ……………...(3)
Substituting (2) and (3) in (1) we get
10−10Δp=6.62×10−344π{10^{ - 10}}\Delta p = \dfrac{{6.62 \times {{10}^{ - 34}}}}{{4\pi }}
Multiplying by 10−10{10^{ - 10}} on both the sides we get
Δp=6.62×10−244πkgms−1\Delta p = \dfrac{{6.62 \times {{10}^{ - 24}}}}{{4\pi }}kgm{s^{ - 1}}
On solving we get
Δp=0.53×10−24kgms−1\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 0.53×10−24kgms−10.53 \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.