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

6.91 x 10^-23 kg m/s

Step-by-step 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-(v^(2))/(c^(2))}}`

`p=\frac{9.1* 10^(-31)* 0.965* 3* 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.

Answer 2

`1.0047* 10^(-21)kg.m/s`

Step-by-step explanation:

Relativistic momentum is given by:

`p=\gamma m u`

where,
`\gamma=\frac{1}{\sqrt{1-(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-((0.965c)^2)/(c^2)}}=(1)/(√(0.2622))=3.81`

`p=3.81* 9.1* 10^(-31)* 0.965* 3* 10^8=1.0047* 10^(-21)kg.m/s`