Question

An electron with a speed of 0.939 is emitted by a supernova, where is the speed of light. What is the magnitude of the momentum of this electron?

Answer 1

Magnitude of momentum = 4.27×10^-22Kg/S

Step-by-step explanation:

Relativistic momentum is given by:

P= mc/(sqrt(1-V^2/C^2)

Po= mo× 0.939c/(sqrt(1-0.939C)^2/C^2)

Po= mo×9.39c/6

Substituting values of no and c

P= (9×10^-31)×9.39×(3×10^8)/6

P= 2.56×10^-21/6

P= 4.27×10^-22Kg/S

Answer 2

Step-by-step explanation:

It is given that,

Speed of electron, v = 0.939 c

c is the speed of light

Mass of electron,
`m=9.1* 10^(-31)\ kg`

The momentum of this electron is given by the product of mass and velocity. It is given that,

p = m v

`p=9.1* 10^(-31)\ kg* 0.939* 3* 10^8`

`p=2.56* 10^(-22)`

So, the momentum of the electron is
`p=2.56* 10^(-22)`. Hence, this is the required solution.