
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
594k+ 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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

The largest wind power cluster is located in the state class 11 social science CBSE

Explain zero factorial class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Which among the following are examples of coming together class 11 social science CBSE

Can anyone list 10 advantages and disadvantages of friction

