Question

Uncertainty in position is twice the uncertainty in momentum, uncertainty in velocity is:
(A) $\sqrt {\dfrac{{\text{h}}}{\pi }}$
(B) $\dfrac{1}{{2{\text{m}}}}\sqrt {\dfrac{{\text{h}}}{\pi }}$
(C) $\dfrac{1}{{2{\text{m}}}}\sqrt {\text{h}}$
(D) $\dfrac{1}{{2\sqrt 2 {\text{m}}}}\sqrt {\dfrac{{\text{h}}}{\pi }}$

Answer 1

Heisenberg uncertainty principle states that it is impossible to simultaneously measure the exact position and exact momentum of a body like an electron. We have been given the relation between the uncertainty in momentum and the uncertainty in the momentum. Using that we will get the uncertainty in velocity.
Formula: $\Delta {\text{x}}\Delta {\text{p}} = \dfrac{{\text{h}}}{{4\pi }}$ and $\Delta {\text{p}} = {\text{m}}\Delta {\text{v}}$ are the mathematical representations of Heisenberg uncertainty principle.

Complete step by step solution:
The Heisenberg uncertainty principle was given by Werner Heisenberg in 1927. It states that it is impossible to simultaneously measure the exact position and exact momentum of a body like an electron. Mathematically representation of Heisenberg uncertainty principle is:
$\Delta {\text{x}}\Delta {\text{p}} \geqslant \dfrac{{\text{h}}}{{4\pi }}{\text{ or }}\Delta {\text{x}}\left( {{\text{m}}\Delta {\text{v}}} \right) \geqslant \dfrac{{\text{h}}}{{4\pi }}$ where $\Delta {\text{x}},\Delta {\text{p}},\Delta {\text{v}}$ are uncertainty in position, uncertainty in momentum and uncertainty in velocity respectively. As momentum is equal to product of mass and velocity, thereby uncertainty in momentum can be represented in terms of uncertainty in velocity as: $\Delta {\text{p}} = {\text{m}}\Delta {\text{v}}$ . Uncertainty in mass is not possible. Value of $\dfrac{{\text{h}}}{{4\pi }}$ is equal to $0.527 \times {10^{ - 34}}{\text{Jsec}}$ . It can be understood in this way simply, as a product of uncertainty in momentum and uncertainty in position is equal to a constant. So, every time we try to approach the accurate momentum of an electron and make uncertainty in momentum less; then uncertainty in position will increase in order to keep their product value to be constant. Thus, it is not possible to accurately and simultaneously predict the exact momentum and position of electrons at a given instant of time.
As according to our question, the uncertainty in position is twice the uncertainty in momentum means $\Delta {\text{x}} = 2\Delta {\text{p}}$ . Then:
$\Delta {\text{x}}\Delta {\text{p}} = \dfrac{{\text{h}}}{{4\pi }}$
$\left( {2\Delta {\text{p}}} \right)\Delta {\text{p}} = \dfrac{{\text{h}}}{{4\pi }}$
$2\Delta {{\text{p}}^2} = \dfrac{{\text{h}}}{{4\pi }}$
$\Delta {\text{p}} = \dfrac{1}{{2\sqrt 2 }}\sqrt {\dfrac{{\text{h}}}{\pi }}$
Putting the value of uncertainty in momentum in the equation $\Delta {\text{p}} = {\text{m}}\Delta {\text{v}}$ , we get:
$\Delta {\text{v}} = \dfrac{1}{{2\sqrt 2 {\text{m}}}}\sqrt {\dfrac{{\text{h}}}{\pi }}$ .
Thus, the correct option is D.

Note:
According to Bohr’s theory which considered electrons as a material particle and therefore its position and momentum can be determined accurately. But according to de-Broglie’s hypothesis, an electron is a form of wave. Thereby exact position and velocity could not be calculated accurately simultaneously at a given time.