Question

In positron-emission tomography (PET) used in medical research and diagnosis, compounds containing unstable nuclei that emit positrons are introduced into the brain, destined for a site of interest in the brain. When a positron is emitted, it goes only a short distance before coming nearly to rest. It forms a bound state with an electron, called "positronium", which is rather similar to a hydrogen atom. The binding energy of positronium is very small compared to the rest energy of an electron. After a short time the positron and electron annihilate. In the annihilation, the positron and the electron disappear, and all of their rest energy goes into two photons (particles of light) which have zero mass; all their energy is kinetic energy. These high energy photons, called "gamma rays", are emitted at nearly 180° to each other. What energy of gamma ray (in MeV, million electron volts) should each of the detectors be made sensitive to? (The mass of an electron or positron is 9 × 10-31 kg. 1 eV = 1.6 × 10-19 joules.

Answer 1

Answer:

Energy of gamma ray = 506250 eV

Explanation:

We are told that the mass of an electron or positron is 9 × 10-31 kg.

This means that their energies will be the same.

Thus; E_e = E_p

Now, since electron and positron annihilate to form gamma(γ) particle, then using work energy principle, we have;

E_γ + E_γ = E_e + E_p

Thus;

2E_γ = E_e + E_p

Since E_e = E_p, we now have;

2E_γ = 2E_e

Thus;

E_γ = E_e

Since all the energy of the electron is converted, then from Einstein's relativity theory, this implies that;

E_e = mc²

c is speed of light = 3 × 10^(8) m/s

And m is given as 9 × 10^(-31) kg

Thus;

E_e = 9 × 10^(-31) × (3 × 10^(8))^(2)

E_e = 810 × 10^(-16) J

Since E_γ = E_e

Thus;

E_γ = 810 × 10^(-16) J

We are given that; 1.6 × 10^(-19) J = 1eV

Thus; 810 × 10^(-16) J gives;

(810 × 10^(-16) × 1)/(1.6 × 10^(-19)) = 506250 eV