Positron annihilated during collision with electron produces — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - The Annihilation Process When a positron ($\ce{e^+}$) collides with an electron ($\ce{e^-}$), the two particles — a particle and its antiparticle — undergo **mutual annihilation**. Their entire rest mass is converted to energy according to Einstein's mass-energy relation. Step 2 - Einstein's Mass-Energy Equivalence Rest mass energy of one electron (or positron): $$E_0 = m_e c^2 = (9.11 \times 10^{-31}\ \text{kg})(3 \times 10^8\ \text{m/s})^2 = 8.19 \times 10^{-14}\ \text{J} = 0.511\ \text{MeV}$$ Total energy converted = $2 \times 0.511\ \text{MeV} = 1.022\ \text{MeV}$ Step 3 - Product of Annihilation Two photons (**γ-rays**), each of energy 0.511 MeV, are produced moving in opposite directions to conserve momentum: $$\ce{e^+ + e^- -> \gamma + \gamma}$$ Step 4 - Evaluate Options - **(A) γ-rays**: Pair annihilation produces two back-to-back γ photons of 0.511 MeV each. **Correct.** - **(B) neutron**: A neutron cannot be produced in pair annihilation (no baryon number involved). Incorrect. - **(C) proton**: Similarly, no baryon number is available. Incorrect. - **(D) α-particle**: Cannot be created; α-particle production requires nucleons. Incorrect. $$\boxed{\text{Answer: A — } \gamma\text{-rays (two photons of 0.511 MeV each)}}$$