Question

Suppose the minimum uncertainty in the position of a

particleis equal to its de Broglie wavelength. If the particle has
anaverage speed of 4.1 105
m/s,what is the minimum uncertainty in its speed?

Answer 1

The minimum uncertainty in its speed is greater than
`6.52* 10^4\ m/s`.

Step-by-step explanation:

It is given that,

Speed of the particle,
`v=4.1* 10^5\ m/s`

We need to find the minimum uncertainty in its speed. It can be calculated using uncertainty principle as :

`\Delta x.\Delta p\ge (h)/(2\pi)`

Since, p = mv

`\Delta x.\Delta (mv)\ge (h)/(2\pi)`

`((h)/(mv)).(m\Delta v)\ge (h)/(2\pi)`

`(\Delta v)/(v)\ge (h)/(2\pi)`

`\Delta v\ge (v)/(2\pi)`

`\Delta v\ge (4.1* 10^5)/(2\pi)`

`\Delta v\ge 6.52* 10^4\ m/s`

So, the minimum uncertainty in its speed is greater than
`6.52* 10^4\ m/s`. Hence, this is the required solution.