Respuesta :
Answer:
The minimum uncertainty in its speed is greater than [tex]6.52\times 10^4\ m/s[/tex].
Explanation:
It is given that,
Speed of the particle, [tex]v=4.1\times 10^5\ m/s[/tex]
We need to find the minimum uncertainty in its speed. It can be calculated using uncertainty principle as :
[tex]\Delta x.\Delta p\ge \dfrac{h}{2\pi}[/tex]
Since, p = mv
[tex]\Delta x.\Delta (mv)\ge \dfrac{h}{2\pi}[/tex]
[tex](\dfrac{h}{mv}).(m\Delta v)\ge \dfrac{h}{2\pi}[/tex]
[tex]\dfrac{\Delta v}{v}\ge \dfrac{h}{2\pi}[/tex]
[tex]\Delta v\ge \dfrac{v}{2\pi}[/tex]
[tex]\Delta v\ge \dfrac{4.1\times 10^5}{2\pi}[/tex]
[tex]\Delta v\ge 6.52\times 10^4\ m/s[/tex]
So, the minimum uncertainty in its speed is greater than [tex]6.52\times 10^4\ m/s[/tex]. Hence, this is the required solution.
