Question

The radionuclide ¹¹C decays according to

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values:

calculate Q and compare it with the maximum energy of the positron emitted

Answer 1

The given nuclear reaction is:

Atomic mass of = 11.011434 u

Atomic mass of

Maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the nucleus is given as:

Where,

m_e = Mass of an electron or positron = 0.000548 u

c = Speed of light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 m_e in the case ofand 5 m_e in the case of.

Hence, equation (1) reduces to:

∴ΔQ = [11.011434 − 11.009305 − 2 × 0.000548] c²

= (0.001033 c²) u

But 1 u = 931.5 Mev/c²

∴ΔQ = 0.001033 × 931.5 ≈ 0.962 MeV

The value of Q is almost comparable to the maximum energy of the emitted positron.