Question

It is desired to create a particle of mass 7920 MeV/c^2 in a head-on collision between a proton and an antiproton (each having a mass of 938.3 MeV/c^2) traveling at the same speed. What peed is necessary for this to occur?

Answer 1

Answer:

$v=0.9714c$

Explanation:

The kinetic energy possessed by  particles will be  

$K.E=\frac{1}{2}Mc^2$

where,

M is the mass of the particle (7920938.3 MeV/c²)

c is the speed of the light

Also,

energy of the proton particle = $\frac{m_pc^2}{\sqrt{1-\frac{v^2}{c^2}}}$

where,

v is the velocity

m_p is the mass of the proton (938.3 MeV/c²)

since the energy is equal

thus,

$\frac{m_pc^2}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{2}Mc^2$

or

$1-\frac{v^2}{c^2}=[\frac{2m_p}{M}]^2$

substituting the values in the above equation, we get

$1-\frac{v^2}{c^2}=[\frac{2\times 938.3 }{7920}]^2$

or

$v=0.9714c$

Hence, the speed necessary for the specified condition to occur is 0.9714 times the speed of the light