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Question: The mass of an electron is \(9.11 \times {10^{ - 31}}\) kg. Planck’s constant is \(6.626 \times {10^...

The mass of an electron is 9.11×10−319.11 \times {10^{ - 31}} kg. Planck’s constant is 6.626×10−346.626 \times {10^{ - 34}} Js then the uncertainty involved in the measurement of velocity within a distance of 0.10.1 angstrom is-
A.5.79×106ms−15.79 \times {10^6}m{s^{ - 1}}
B.5.79×107ms−15.79 \times {10^7}m{s^{ - 1}}
C.5.79×108ms−15.79 \times {10^8}m{s^{ - 1}}
D.5.79×105ms−15.79 \times {10^5}m{s^{ - 1}}

Explanation

Solution

Heisenberg’s uncertainty principle is given by the formula-
Δx.Δp⩾h4π\Delta x.\Delta p \geqslant \dfrac{h}{{4\pi }} where Δx\Delta x is uncertainty in position, Δp\Delta p is uncertainty in momentum, and h is Planck’s constant. Also we know that the formula Δp=mΔv\Delta p = m\Delta v where m is the mass of a particle and Δv\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=9.11×10−319.11 \times {10^{ - 31}} kg
Planck’s constant h=6.626×10−346.626 \times {10^{ - 34}} Js
Distance or uncertainty in positionΔx\Delta x=0.10.1 angstrom=10−10m = {10^{ - 10}}m
Now we know that Heisenberg’s uncertainty principle is given by the formula-
⇒\Rightarrow Δx.Δp⩾h4π\Delta x.\Delta p \geqslant \dfrac{h}{{4\pi }} --- (i)
where Δx\Delta x is uncertainty in position, Δp\Delta p is uncertainty in momentum, and h is Planck’s constant.
Also, Δp=mΔv\Delta p = m\Delta v-- (ii)
where m is the mass of the particle and Δv\Delta v is uncertainty in velocity.
From eq. (i) and (ii) we can write,
⇒Δx.mΔv⩾h4π\Rightarrow \Delta x.m\Delta v \geqslant \dfrac{h}{{4\pi }}
We can rearrange it and write as-
⇒Δx.Δv⩾h4πm\Rightarrow \Delta x.\Delta v \geqslant \dfrac{h}{{4\pi m}}
Now putting the given values in the formula, we get-
⇒Δv.(10−10)⩾6.626×10−344×227×9.1×10−31\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,
⇒Δv⩾6.626×10−34×74×22×9.1×10−31×10−10\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,
⇒Δv⩾6.626×10−34×788×9.1×10−41\Rightarrow \Delta v \geqslant \dfrac{{6.626 \times {{10}^{ - 34}} \times 7}}{{88 \times 9.1 \times {{10}^{ - 41}}}}
⇒Δv⩾46.382×107800.8\Rightarrow \Delta v \geqslant \dfrac{{46.382 \times {{10}^7}}}{{800.8}}
On division we get,
⇒Δv⩾0.0579195×107\Rightarrow \Delta v \geqslant 0.0579195 \times {10^7}
On multiplying we get,
⇒Δv⩾5.79×105ms−1\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 0.10.1 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-
ΔE.Δt⩾h4π\Delta E.\Delta t \geqslant \dfrac{h}{{4\pi }} where ΔE\Delta E is uncertainty in energy and Δt\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.