Question

The location of a particle is measured with an uncertainty of 0.13 nm. One tries to simultaneously measure the velocity of this particle. What is the minimum uncertainty in the velocity measurement. The mass of the particle is 1.745×10-27 kg

Answer 1

The minimum uncertainty in the velocity is 232.57 m/s.

Step-by-step explanation:

Given that,

Location of a particle with uncertainty = 0.13 nm

Mass of particle
`m=1,745*10^(-27)\ kg`

We need to calculate the minimum uncertainty in the velocity

Using heisenberg's uncertainty principle,

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

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

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

Put the value into the formula

`\Delta v=(6.63*10^(-34))/(4\pi* 1.745*10^(-27)*0.13*10^(-9))`

`\Delta v= 232.57\ m/s`

Hence, The minimum uncertainty in the velocity is 232.57 m/s.