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Question: A dust particle has mass equal to \[{{10}^{-11}}g\], diameter \[{{10}^{-4}}cm\] and velocity \[{{10}...

A dust particle has mass equal to 10−11g{{10}^{-11}}g, diameter 10−4cm{{10}^{-4}}cm and velocity 10−4 cm sec⁡−1{{10}^{-4}}\,cm\,{{\sec }^{-1}}. The error in measurement of velocity is 0.1%. What will be the uncertainty in its position?

Explanation

Solution

Hint: Heisenberg’s uncertainty principle is a key principle in quantum mechanics. It states that if we have information about where a particle is located (the uncertainty of position is small), we cannot say anything about its momentum (the uncertainty of momentum is large), and vice versa.

Complete step-by-step answer:
The Heisenberg uncertainty principle states that the position and the velocity of an object, both cannot be measured exactly, at the same time, even in theory. The concepts of exact position and exact velocity together, have no meaning in nature.
The product of the uncertainty in position of a particle and the uncertainty in its momentum can never be less than one-fourth of the Planck constant:

ΔpΔx≥h4π\Delta p\Delta x\ge \dfrac{h}{4\pi }, (i)
where h= Planck's constant = 6.62×10−34 Jsec⁡6.62\times {{10}^{-34}}\,J\sec
Δp\Delta p = change in momentum
Δx\Delta x= uncertainty in position
We know that momentum (p) is also written in terms of mass (m) and velocity (v) as p = mv.
Therefore, Δp\Delta p can also be written as Δp=mΔv\Delta p=m\Delta v.

We have been given in the question that the:
Velocity of the particle,v=10−4 cm sec⁡−1v={{10}^{-4}}\,cm\,{{\sec }^{-1}}
Diameter of the particle, d=10−4cmd={{10}^{-4}}cm
Mass of the particle, m=10−11gm={{10}^{-11}}g
By using this uncertainty relation, we can calculate the uncertainty in position, if uncertainty in momentum is given.
Δv=0.1×10−4100=1×10−7\Delta v=\dfrac{0.1\times {{10}^{-4}}}{100}=1\times {{10}^{-7}}
By substituting the momentum in the equation (i) we can write,

mΔvΔx≥h4πm\Delta v\Delta x\ge \dfrac{h}{4\pi }
On rearranging we get;
Δx=h4πm×Δv\Delta x=\dfrac{h}{4\pi m\times \Delta v}
On substituting the values as provided in the question, we get;

& \Delta x\,=\dfrac{6.626\times {{10}^{-27}}}{4\times 3.14\times {{10}^{-11}}\times {{10}^{-7}}} \\\ & \,\,\,\,\,\,\,=0.527\times {{10}^{-9}}cm \\\ \end{aligned}$$ Therefore, the uncertainty in position is calculated as $$0.527\times {{10}^{-9}}cm$$. Note: The uncertainty principle formally limits the precision to which two complementary observables can be measured and establishes that observables are not independent of the observer. It also establishes that phenomena can take on a range of values rather than a single, exact value.