Question

Suppose the minimum uncertainty in the position of a

Answer 1

Answer:

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

Explanation:

It is given that,

Speed of the particle, $v=4.1\times 10^5\ m/s$

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

$\Delta x.\Delta p\ge \dfrac{h}{2\pi}$

Since, p = mv

$\Delta x.\Delta (mv)\ge \dfrac{h}{2\pi}$

$(\dfrac{h}{mv}).(m\Delta v)\ge \dfrac{h}{2\pi}$

$\dfrac{\Delta v}{v}\ge \dfrac{h}{2\pi}$

$\Delta v\ge \dfrac{v}{2\pi}$

$\Delta v\ge \dfrac{4.1\times 10^5}{2\pi}$

$\Delta v\ge 6.52\times 10^4\ m/s$

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