Question

For a nuclear reaction A + b →  c + d, the Q-value is defined as Q = [m_A + m_b - m_c - m_d]c². A radioactive nuclei 11/6 C decays in following way 11/6 C→ 11/5 +e⁺ v; T_1/2 = 20.3 minutes. the maximum energy emitted by positron is 0.960 MeV. Following values are. given. m((11/6)C) = 11.011434u,m (11B/5) = 11.009305u,m_e = 0.000548 μ.

Evaluate Q-value of this reaction and compare it with maximum Energy Emitted by Positron.

Answer 1

Equation of = decay is following-

This energy is comparable. to the maximum energy emitted by positron 1033 x 931 MeV.

∴ Q = 0.96 Mev.