Conceptual Question on Heisenberg Uncertainty Principle. #5

 QUESTION:

Suppose the speed of electrons present in the first shell of an atom is 50% of the speed of light. If the uncertainty in velocity is 1000\rm \ m/s, what is the uncertainty in the position of this electron?

 SOLUTION:

Let’s suppose:

  • m_e is the mass of the electron, and

  • V is the speed of the electron.

Given:

  • V = \dfrac{c}{2}, and

  • \Delta V = 1000 \ m/s.

To find:

  • \Delta x

We know the value:

  • h = 6.626 \times 10^{ -34}\rm \  J \cdot s, and

  • Rest mass of electron, m_e = 9.11\times 10^{-31}\rm \ kg.

Here the speed of the electron is comparable to the speed of light so the mass of the electron at this speed will be quite more than the rest mass of the electron. So, let’s calculate it.

The relativistic mass of the electron when its speed is 50% of the speed of light is given by

\begin{aligned}m &= m_e \dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\&= (9.11\times 10^{-31}\rm \ kg) \times \dfrac{1}{\sqrt{1-\frac{1}{4}}}\\&= 12.147 \times 10^{-31}\rm \ kg \\\end{aligned}

The uncertainty in the momentum is;

\begin{aligned}\Delta P &= m\Delta V\\&= (12.147 \times 10^{-31}\rm \ kg )\times (1000 \ m/s )\\&= 12.147 \times 10^{-27}\rm \ kg \cdot \ m/s\\\end{aligned}

From the Heisenberg uncertainty principle, the uncertainty in position and the uncertainty in momentum are related by an equation;

\begin{aligned}\Delta x \times \Delta P &\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta x &\geq \dfrac{h}{4\pi\Delta P}\\ &= \dfrac{6.626 \times 10^{ -34}\rm \ : J \cdot s}{4\pi \times (12.147 \times 10^{-27}\rm \ kg \cdot \ m/s)}\\&= 4.34 \times 10^{ -9} \rm \ m\\\end{aligned}


Therefore, the minimum uncertainty in the particle’s position is

\begin{aligned}\Delta x &= 4.34 \times 10^{ -9} \rm \ m \\\end{aligned}

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