Question

The mass of an electron is $9.11 \times {10^{ - 31}}$ kg. Planck’s constant is $6.626 \times {10^{ - 34}}$ Js then the uncertainty involved in the measurement of velocity within a distance of $0.1$ angstrom is-
A.$5.79 \times {10^6}m{s^{ - 1}}$
B.$5.79 \times {10^7}m{s^{ - 1}}$
C.$5.79 \times {10^8}m{s^{ - 1}}$
D.$5.79 \times {10^5}m{s^{ - 1}}$

Answer 1

Hint: Heisenberg’s uncertainty principle is given by the formula-
$\Delta x.\Delta p \geqslant \dfrac{h}{{4\pi }}$ where $\Delta x$ is uncertainty in position, $\Delta p$ is uncertainty in momentum, and h is Planck’s constant. Also we know that the formula \[\Delta p = m\Delta v\] where m is the mass of a particle and $\Delta v$ is uncertainty in velocity. Use these two formulas to find the certainty of velocity.

Complete step by step answer:
Given the mass of electron m=$9.11 \times {10^{ - 31}}$ kg
Planck’s constant h=$6.626 \times {10^{ - 34}}$ Js
Distance or uncertainty in position$\Delta x$=$0.1$ angstrom$ = {10^{ - 10}}m$
Now we know that Heisenberg’s uncertainty principle is given by the formula-
$ \Rightarrow $ $\Delta x.\Delta p \geqslant \dfrac{h}{{4\pi }}$ --- (i)
where $\Delta x$ is uncertainty in position, $\Delta p$ is uncertainty in momentum, and h is Planck’s constant.
Also, \[\Delta p = m\Delta v\]-- (ii)
 where m is the mass of the particle and $\Delta v$ is uncertainty in velocity.
From eq. (i) and (ii) we can write,
$ \Rightarrow \Delta x.m\Delta v \geqslant \dfrac{h}{{4\pi }}$
We can rearrange it and write as-
$ \Rightarrow \Delta x.\Delta v \geqslant \dfrac{h}{{4\pi m}}$
Now putting the given values in the formula, we get-
$ \Rightarrow \Delta v.\left( {{{10}^{ - 10}}} \right) \geqslant \dfrac{{6.626 \times {{10}^{ - 34}}}}{{4 \times \dfrac{{22}}{7} \times 9.1 \times {{10}^{ - 31}}}}$
On adjusting we get,
$ \Rightarrow \Delta v \geqslant \dfrac{{6.626 \times {{10}^{ - 34}} \times 7}}{{4 \times 22 \times 9.1 \times {{10}^{ - 31}} \times {{10}^{ - 10}}}}$
On solving we get,
$ \Rightarrow \Delta v \geqslant \dfrac{{6.626 \times {{10}^{ - 34}} \times 7}}{{88 \times 9.1 \times {{10}^{ - 41}}}}$
$ \Rightarrow \Delta v \geqslant \dfrac{{46.382 \times {{10}^7}}}{{800.8}}$
On division we get,
$ \Rightarrow \Delta v \geqslant 0.0579195 \times {10^7}$
On multiplying we get,
$ \Rightarrow \Delta v \geqslant 5.79 \times {10^5}m{s^{ - 1}}$
This is the value of uncertainty involved in measurement of velocity of electrons within the distance of $0.1$ angstrom.

Hence the correct option is D.

Note:
Heisenberg’s uncertainty principle tells us that position and momentum of a particle cannot be simultaneously measured with high precision. It is important for microscopic particles. It is also given by-
$\Delta E.\Delta t \geqslant \dfrac{h}{{4\pi }}$ where $\Delta E$ is uncertainty in energy and $\Delta t$ is uncertainty in time.
Here it tells us that the energy of a photon is less than the energy needed to change the position and velocity of bigger bodies when it collides with them.