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

Answer:

0.572 MeV

2.467 MeV

Explanation:

c = Speed of light = $3\times 10^8\ m/s$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

v = Final velocity

u = Initial velocity

Kinetic energy is given by

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

The energy required is 0.572 MeV

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

The energy required is 2.467 MeV