
The size of a microscopic particle is 1 micron and its mass is $6 \times {10^{ - 13}}$g. If its position may be measured to within 0.1% of its size, the uncertainty in velocity (in $c{m^{ - 1}}$) is approximately ?
a.) $\dfrac{{{{10}^{ - 7}}}}{{4\Pi }}$
b.) $\dfrac{{{{10}^{ - 5}}}}{{4\Pi }}$
c.) ${10^{ - 5}}$
d.) ${10^{ - 8}}$
Answer
570.9k+ views
Hint: This question is based on Heisenberg’s uncertainty principle which can be mathematically written as -
$\Delta x \times \Delta p \geqslant \dfrac{h}{{2\Pi }}$
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 -
$\Delta x \times \Delta p \geqslant \dfrac{h}{{2\Pi }}$
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 -
$\Delta x \times m\Delta v \geqslant \dfrac{h}{{2\Pi }}$
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) = $6 \times {10^{ - 13}}$g
Δx = 0.1% of size of the particle
Δx = $\dfrac{{0.1}}{{100}} \times {10^{ - 4}}$
Δx = ${10^{ - 7}}cm$
To find :
Uncertainty in velocity (in $c{m^{ - 1}}$) -
Now, filling all the values in the above formula, we get,
${10^{ - 7}} \times 6 \times {10^{ - 13}}\Delta v$ = $\dfrac{h}{{4\Pi }}$
On solving the above equation, we get -
Δv = $\dfrac{{6.626 \times {{10}^{ - 34}}}}{{4 \times 3.14 \times {{10}^{ - 7}} \times 6 \times {{10}^{ - 13}}}}$
Δv = $0.276 \times {10^{ - 14}}c{m^{ - 1}}$
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.
$\Delta x \times \Delta p \geqslant \dfrac{h}{{2\Pi }}$
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 -
$\Delta x \times \Delta p \geqslant \dfrac{h}{{2\Pi }}$
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 -
$\Delta x \times m\Delta v \geqslant \dfrac{h}{{2\Pi }}$
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) = $6 \times {10^{ - 13}}$g
Δx = 0.1% of size of the particle
Δx = $\dfrac{{0.1}}{{100}} \times {10^{ - 4}}$
Δx = ${10^{ - 7}}cm$
To find :
Uncertainty in velocity (in $c{m^{ - 1}}$) -
Now, filling all the values in the above formula, we get,
${10^{ - 7}} \times 6 \times {10^{ - 13}}\Delta v$ = $\dfrac{h}{{4\Pi }}$
On solving the above equation, we get -
Δv = $\dfrac{{6.626 \times {{10}^{ - 34}}}}{{4 \times 3.14 \times {{10}^{ - 7}} \times 6 \times {{10}^{ - 13}}}}$
Δv = $0.276 \times {10^{ - 14}}c{m^{ - 1}}$
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.
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