The total energy of a particle is 4.11 times its rest energy. The mass of the pa

Question

The total energy of a particle is 4.11 times its rest energy. The mass of the particle is 5.6 × 10-27 kg. Find the particle's rest energy. The speed of light is 2.99792 x 103 m/s and 1 J 6.242 x 1012 MeV Answer in units ofMeV 013 (part 2 of 4) 10.0 points With what speed is the particle moving? Answer in units of m/S. 014 (part 3 of 4) 10.0 points Determine the kinetic energy of the particle. Answer in units ofMeV 015 (part 4 of 4) 10.0 points What is the particle's momentum? Answer in units of MeV/c

Explanation / Answer

Part 1:

Rest energy of a particle is given by:

E0 = m*c^2

Using given values:

E0 = 5.6*10^-27 kg*(2.99792*10^8)^2 m/sec

E0 = 5.03301362*10^-10 J

1 J = 6.242*10^12 MeV

E0 = 5.03301362*10^-10*6.242*10^12

E0 = 3141.6071 MeV

Part 2

Given that

E = 4.11*E0

E = m*c^2/sqrt (1 - v^2/c^2) = E0/sqrt (1 - v^2/c^2)

4.11*E0 = E0/sqrt (1 - v^2/c^2)

1 - v^2/c^2 = 1/4.11^2

v^2 = c^2*(1 - 1/4.11^2)

v = c*sqrt (1 - 1/4.11^2)

v = 2.99792*10^8*sqrt (1 - 1/4.11^2)

v = 290782899 = 2.90783*10^8 m/sec

Part 3.

KE of particle will be

KE = E - mc^2

KE = 4.11*E0 - E0

KE = 3.11*3141.6071 MeV

KE = 9770.3981 MeV

Part 4

Momentum is given by:

P = m*c*sqrt (n^2 - 1)

n = 4.11

P = (mc^2/c)*sqrt (n^2 - 1) = (E0/c)*sqrt (n^2 - 1)

P = (3141.6071 MeV/c)*sqrt (4.11^2 - 1)

P = 12523.9843 MeV/c

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