Question

Determine the energy required to accelerate an electron between each of the following speeds.(a) 0.500c to 0.898c(b) 0.898c to 0.990c

Answer 1

0.572 MeV

2.467 MeV

Step-by-step explanation:

c = Speed of light =
`3* 10^8\ m/s`

m = Mass of electron =
`9.11* 10^(-31)\ kg`

v = Final velocity

u = Initial velocity

Kinetic energy is given by

`E=mc^2\left[\frac{1}{\sqrt{1-(v^2)/(c^2)}}-\frac{1}{\sqrt{1-(u^2)/(c^2)}}\right]\\\Rightarrow E=9.11* 10^(-31)* (3* 10^8)^2\left[\frac{1}{\sqrt{1-(0.898^2c^2)/(c^2)}}-\frac{1}{\sqrt{1-(0.5^2c^2)/(c^2)}}\right]\\\Rightarrow E=9.16689* 10^(-14)\ J\\\Rightarrow E=(9.16689* 10^(-14))/(1.6* 10^(-13))\\\Rightarrow E=0.572\ MeV`

The energy required is 0.572 MeV

`E=mc^2\left[\frac{1}{\sqrt{1-(v^2)/(c^2)}}-\frac{1}{\sqrt{1-(u^2)/(c^2)}}\right]\\\Rightarrow E=9.11* 10^(-31)* (3* 10^8)^2\left[\frac{1}{\sqrt{1-(0.99^2c^2)/(c^2)}}-\frac{1}{\sqrt{1-(0.898^2c^2)/(c^2)}}\right]\\\Rightarrow E=3.94869* 10^(-13)\ J\\\Rightarrow E=(3.94869* 10^(-13))/(1.6* 10^(-13))\\\Rightarrow E=2.467\ MeV`

The energy required is 2.467 MeV