Question

If the position of the electron is measured within an accuracy of ±0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is$\dfrac{h}{4\pi m}\times 0.05nm$, is there any problem in defining this value?

Answer 1

The question is related to Heisenberg’s uncertainty principle. According to the uncertainty principle there will always be an error while simultaneously measuring the position and momentum of a body in the quantum or microscopic world. From the mathematical expression of this uncertainty principle we would be able to find the uncertainty in the momentum of an electron.

Complete step by step answer:
- Let’s start with the concept of Heisenberg’s uncertainty principle. The principle states that we cannot measure the momentum (p) and the position (x) of a particle simultaneously with absolute precision. That is the more accurately we know one of these values (momentum or position), the less accurately we know the other.
- Multiplying together the errors in the measurements of the values of the position and momentum, will give rise to a number greater than or equal to half of a constant called "h-bar" and this quantity is equal to Planck's constant (h) divided by 2π.
- The mathematical representation of this law can be given as follows
$\Delta x.\Delta p\ge \dfrac{h}{4\pi }$
Where $\Delta x$ is the uncertainty in the position of electron
$\Delta p$ is the uncertainty in the momentum of electron
h is the Planck’s constant ( $6.626\times {{10}^{-34}}Js$)
Since we are asked to find the uncertainty in the momentum of the electron we can rearrange the above equation as follows,
$\Delta p=\dfrac{1}{\Delta x}.\dfrac{h}{4\pi }$
Substituting the given values in the equation we get,
$\Delta p=\dfrac{1}{0.002}nm\times \dfrac{6.626\times {{10}^{-34}}Js}{4\times 3.14}$
As we know one nanometer is ${{10}^{-9}}$m. Hence we can write 0.002 nm as $2\times {{10}^{-12}}m$. Substitute this in the above equation
$\Delta p=\dfrac{1}{2\times {{10}^{-12}}m}nm\times \dfrac{6.626\times {{10}^{-34}}Js}{4\times 3.14}$
$\Delta p=2.637\times {{10}^{-23}}kgm{{s}^{-1}}$

Therefore the uncertainty in the momentum of the electron is $2.637\times {{10}^{-23}}kgm{{s}^{-1}}$.
$Actual\text{ }momentum=\dfrac{h}{4\pi m\times 0.05nm}$
$=\dfrac{6.626\times {{10}^{-34}}Js}{4\times 3.14\times 5.0\times {{10}^{-11}}m}$
$=1.055\times {{10}^{-24}}kgm{{s}^{-1}}$
We can't take exact value of momentum since the magnitude of the actual momentum is smaller than the uncertainty and as a result, it will violate Heisenberg’s uncertainty principle and the value $\dfrac{h}{4\pi m}\times 0.05nm$ cannot be defined.

Note: It should be noted that the uncertainty principle is the reason for the failure of Bohr’s atomic model. According to Bohr the electron revolves around the nuclei in the fixed orbits with fixed velocity. Thus he said that the momentum and position of the electron can be found accurately, which is in strict disparity with the uncertainty principle.