Question

The position of an electron is measured within an uncertainty of 0.100 nm. What will be its minimum position uncertainty 2.00 s later? {3.32 x 106 m}

Answer 1

Answer:

Minimum uncertainty in position is $\Delta x= 1157808.48\ m$

Explanation:

It is given that,

Uncertainty in the position of an electron, $\Delta x=0.1\ nm=0.1\times 10^{-9}\ m$

According to uncertainty principle,

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

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

$\Delta v\geq \dfrac{h}{4\pi \times \Delta x\times m}$

$\Delta v\geq \dfrac{6.62\times 10^{-34}\ J-s}{4\pi \times 0.1\times 10^{-9}\ m\times 9.1\times 10^{-31}\ kg}$

$\Delta v\geq 578904.24\ m/s$

Let $\Delta x$ is the uncertainty in position after 2 seconds such that,

$\Delta x=\Delta v\times t$

$\Delta x=578904.24\ m/s\times 2\ s$

$\Delta x= 1157808.48\ m$

or

$\Delta x= 1.15\times 10^6\ m$

Hence, this is the required solution.