Question

According to the Heisenberg uncertainty principle, if the uncertainty in the speed of an electron is 3.5 × 103 m/s, the uncertainty in its position (in m) is at least:

Answer 1

Explanation:

It is given that,

Uncertainty in the speed of an electron, $\Delta v=3.5\times 10^3\ m/s$

According to Heisenberg uncertainty principle,

$\Delta x.\Delta p=\dfrac{h}{4\pi}$

$\Delta x$ is the uncertainty in the position of an electron

Since, $\Delta p=m\Delta v$

$\Delta x=\dfrac{h}{4\pi.m \Delta v}$

$\Delta x=\dfrac{6.6\times 10^{-34}}{4\pi\times 9.1\times 10^{-31}\times 3.5\times 10^3}$

$\Delta x=1.64\times 10^{-8}\ m$

So, the uncertainty in its position is $1.64\times 10^{-8}\ m$. Hence, this is the required solution.