Respuesta :
Answer:
The minimum kinetic energy of the particle, [tex]K_{min}=0.808 MeV [/tex]
Explanation:
In this question we have given,
[tex]\delta p=2.1\times 10^{-20}kg.m.s^{-1}[/tex]
mass, m=[tex]1.7\times 10^{-27}kg[/tex]
we have to find the minimum kinetic energy of the particle, [tex]K_{min}[/tex]=?
We know that momentum and velocity are related by following formula,
[tex]p=m\time v[/tex]
Therefore
[tex]2.1\times 10^{-20}kg.m.s^{-1}=1.7\times 10^{-27} kg\times v[/tex]
[tex]v=\frac{2.1\times 10^{-20}kg.m.s^{-1}}{1.7\times 10^{-27} kg}[/tex]
[tex]v=1.235\times 10^7 ms^{-1}[/tex]
Now we know that kinetic energy is given as
[tex]K.E=\frac{mv^2}{2}[/tex]
Therefore minimum kinetic energy is given as
[tex]K_{min}=\frac{1.7\times 10^{-27}kg\times (1.235\times 10^7 ms^{-1})^2}{2}[/tex]
[tex]K_{min}=1.295\times 10^{-13} joule[/tex]
[tex]K_{min}=1.295\times 10^{-13}\times 6.242\times 10^18 eV [/tex]
[tex]K_{min}=0.808\times 10^6 eV [/tex]
[tex]K_{min}=0.808 MeV [/tex]
The minimum kinetic energy of the particle[tex]K_{min}=0.808 MeV [/tex]
