Question

What is the linear momentum of 25 gev proton assuming that rest mass energy of the proton is 1 gev?

Answer 1

total energy of proton is given by, $E^2=P^2c^2+m_0^2c^4$

where P is linear momentum, $m_0$ is rest mass and c is speed of light.

a/c to Enstien's relativity theorem,

rest mass energy, $=m_0c^2$

so, $m_0^2c^4=(m_0c^2)=(1geV)^2$

we know, 1geV = 10^9 eV = 1.6 × 10^-10 J

c = 3 × 10^8 m/s .

so, Total energy , (25eV)²= P²(3 × 10^8)² + (1geV)²

or, (25 × 1.6 × 10^-10)² = P² × 9 × 10^16 + (1.6 × 10^-10)²

or, (624 × 1.6 × 10^-10)² = P² × 9 × 10^16

or, 624 × (1.6² × 10^-10)²/9 × 10^16 = P²

or, P = √(624) × 1.6 × 10^-10/3 × 10^8

or, P = 13.32 × 10^-18 Ns

hence, linear momentum of proton is 13.32 × 10^-18 Ns