TOPIC 7 OPTIONS H • The first term (2mc2) represents the mass of both particles.
• The second and third terms (Ek1 and Ek2) are the kinetic energies just before the collision.
• The term on the right-hand side of the equation represents the creation of two photons (2hf). (Each photon has an equal share of energy and hence is of equal frequency.)
The equation indicates that hf must be at least equal to mc2. Since we know that 1.02 MeV is calculated from E = mc2 using two times the mass of an electron, each photon produced here will be 0.51 MeV; so the photons produced correspond to gamma-ray radiation. (This is only in the case of an electron-positron annihilation.)
NOTE
Momenta is the plural of momentum.
If the kinetic energies of the two initial particles were both small, the total momentum before the collision would be close to zero. Because momentum must be conserved, the momentum after annihilation must also be approximately zero. The only way that this can happen is for the two photons to be emitted in opposite directions such that their individual momenta cancel.
QUESTIONS AND ANSWERS h = 6.6 × 10–34 J s c = 3 × 108 m s–1 Mass of proton = 1.672 998 × 10–27 kg
4. Calculate the minimum frequency photon that can cause pair production of two particles equivalent to the mass of a proton and antiproton, each with a velocity of 2.8 × 108 m s−1 after production.
Solution hf = 2(mc2) + Ek1 + Ek2
(6.6 × 10−34)(f) = 2(1.672998 × 10−27) (3 × 108)2 + 2(Ek) (Kinetic energy will be divided evenly as the mass of each particle is identical.)
In order to insert the correct kinetic energy in this equation, we need to calculate it from velocity. Kinetic energy component of each particle: Ek = 1
