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

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

Step-by-step explanation:

It is given that,

Uncertainty in the position of an electron,
`\Delta x=0.1\ nm=0.1* 10^(-9)\ m`

According to uncertainty principle,

`\Delta x.\Delta p\geq (h)/(4\pi)`

`\Delta x.m\Delta v\geq (h)/(4\pi)`

`\Delta v\geq (h)/(4\pi * \Delta x* m)`

`\Delta v\geq (6.62* 10^(-34)\ J-s)/(4\pi * 0.1* 10^(-9)\ m* 9.1* 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* t`

`\Delta x=578904.24\ m/s* 2\ s`

`\Delta x= 1157808.48\ m`

or

`\Delta x= 1.15* 10^6\ m`

Hence, this is the required solution.