Question

An electron with a speed of 0.965 c is emitted by a supernova, where c is the speed of light. What is the magnitude of the momentum of this electron?

Answer 1

Answer:

$1.0047\times 10^{-21}kg.m/s$

Explanation:

Relativistic momentum is given by:

$p=\gamma m u$

where, $\gamma=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$

Speed of the electron, u = 0.965 c (given)

Mass of the electron, m = 9.1 ×10⁻³¹ kg

$\gamma=\frac{1}{\sqrt{1-\frac{(0.965c)^2}{c^2}}}=\frac{1}{\sqrt{0.2622}}=3.81$

$p=3.81\times 9.1\times 10^{-31}\times 0.965\times 3\times 10^8=1.0047\times 10^{-21}kg.m/s$

Answer 2

Answer:

6.91 x 10^-23 kg m/s

Explanation:

velocity, v = 0.965 c

mass of electron, m = 9.1 x 10^-31 kg

The formula for the momentum is given by

$p=\frac{mv}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$p=\frac{9.1\times 10^{-31}\times 0.965\times 3\times 10^{8}}{\sqrt{1-0.965^{2}}}$

p = 6.91 x 10^-23 kg m/s

Thus, the momentum of electron is 6.91 x 10^-23 kg m/s.