Question

A particle has a rest mass of 6.55×10−27 kg and a momentum of 4.34×10−18 kg⋅m/s . Determine the total relativistic energy of the particle

Answer 1

Momentum of a particle is given by:

P = m0*v*

m0 = rest mass = 6.55*10^-27 kg

P = momentum = 4.34*10^-18 kg-m/s

= Lorentz factor = 1/sqrt (1 - v^2/c^2)

v = speed of particle

So,

P = m0*v/sqrt (1 - v^2/c^2)

After re-arranging above equation:

P^2*(1 - v^2/c^2) = m0^2*v^2

P^2 = m0^2*v^2 + P^2*v^2/c^2

v = p*c/sqrt (m0^2*c^2 + P^2)

Using given values:

v = 4.34*10^-18*3*10^8/sqrt ((6.55*10^-27)^2*(3*10^8)^2 + (4.34*10^-18)^2)

v = 2.733*10^8 m/sec

Now total relativistic energy of a particle is given by:

KEtot = m0*c^2*

KEtot = 6.55*10^-27*(3*10^8)^2/sqrt (1 - (2.733*10^8/(3*10^8))^2)

KEtot = 1.43*10^-9 J

Method 2:

Use this direction relation between total relativistic energy and rest mass

KEtot = sqrt (m0^2*c^4 + p^2*c^2)^2

KEtot = sqrt ((6.55*10^-27)^2*(3*10^8)^4 + (3*10^8)^2*(4.34*10^-18)^2)

KEtot = 1.43*10^-9 J

Please Upvote.