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Question: The size of a microscopic particle is 1 micron and its mass is \(6 \times {10^{ - 13}}\)g. If its po...

The size of a microscopic particle is 1 micron and its mass is 6×10−136 \times {10^{ - 13}}g. If its position may be measured to within 0.1% of its size, the uncertainty in velocity (in cm−1c{m^{ - 1}}) is approximately ?
a.) 10−74Π\dfrac{{{{10}^{ - 7}}}}{{4\Pi }}
b.) 10−54Π\dfrac{{{{10}^{ - 5}}}}{{4\Pi }}
c.) 10−5{10^{ - 5}}
d.) 10−8{10^{ - 8}}

Explanation

Solution

This question is based on Heisenberg’s uncertainty principle which can be mathematically written as -
Δx×Δp⩾h2Π\Delta x \times \Delta p \geqslant \dfrac{h}{{2\Pi }}
Where Δx is the uncertainty in position
Δp is the change in momentum
H is the Planck's constant
Further, the moment is the product of the mass of the moving object and the velocity with which it is moving.

Complete step by step answer :
This question is based on Heisenberg’s uncertainty principle which states that it is impossible to measure the position and momentum of a body simultaneously with absolute precision.
Mathematically, it can be written as -
Δx×Δp⩾h2Π\Delta x \times \Delta p \geqslant \dfrac{h}{{2\Pi }}
Where Δx is the uncertainty in position
Δp is the change in momentum
H is the Planck's constant
On filling the value of p = mv
We can write it as -
Δx×mΔv⩾h2Π\Delta x \times m\Delta v \geqslant \dfrac{h}{{2\Pi }}
Where m is the mass of the moving object
And v is the velocity of the object.
Now, let us see the values given to us and what we need to find out.
Given :
Size of a microscopic particle = 1 micron
Mass of the microscopic particle (m) = 6×10−136 \times {10^{ - 13}}g
Δx = 0.1% of size of the particle
Δx = 0.1100×10−4\dfrac{{0.1}}{{100}} \times {10^{ - 4}}
Δx = 10−7cm{10^{ - 7}}cm
To find :
Uncertainty in velocity (in cm−1c{m^{ - 1}}) -
Now, filling all the values in the above formula, we get,
10−7×6×10−13Δv{10^{ - 7}} \times 6 \times {10^{ - 13}}\Delta v = h4Π\dfrac{h}{{4\Pi }}
On solving the above equation, we get -
Δv = 6.626×10−344×3.14×10−7×6×10−13\dfrac{{6.626 \times {{10}^{ - 34}}}}{{4 \times 3.14 \times {{10}^{ - 7}} \times 6 \times {{10}^{ - 13}}}}
Δv = 0.276×10−14cm−10.276 \times {10^{ - 14}}c{m^{ - 1}}
If we see the above options, then none of these matches are answered.

So, none of the options given is correct.

Note: The momentum of a moving object is the product of the mass of the moving object and the velocity with which it is moving. Heisenberg’s uncertainty principle is applicable to only microscopic particles.