Question: Assertion (A) : A free neutron is not a stable particle. Reason (R) : It decays into a proton and an electron.
Step 1 (Governing Law & Reaction Equation): [0.5 Mark]
A free neutron is unstable because its mass is greater than the sum of the masses of a proton and an electron. It undergoes beta-minus ($\beta^-$) decay to achieve a lower energy state. The nuclear reaction is represented as:
$$n \rightarrow p + e^- + \bar{\nu}_e$$Where $n$ is the neutron, $p$ is the proton, $e^-$ is the electron (beta particle), and $\bar{\nu}_e$ is the electron antineutrino.
Step 2 (Step-by-Step Logical Deduction): [0.25 Mark]
- Assertion (A): A free neutron has a mean life of approximately $880$ seconds, confirming it is unstable outside the nucleus.
- Reason (R): The decay process involves the conversion of a down quark into an up quark, resulting in the emission of a proton, an electron, and an antineutrino to conserve charge, lepton number, and energy.
Step 3 (Final Concluded Answer): [0.25 Mark]
Correct Option: Assertion (A) is true, but Reason (R) is false.
Justification: While the neutron does decay into a proton and an electron, the decay process must also include an electron antineutrino ($\bar{\nu}_e$) to satisfy the law of conservation of lepton number. Therefore, the statement in Reason (R) is incomplete and scientifically inaccurate.
💡 EL10 Concept Insight: Think of a free neutron like a spinning top that is slightly wobbly; it can't stay that way forever, so it "breaks" into smaller, more stable pieces (a proton and an electron) plus a tiny, invisible particle called an antineutrino to keep the physics laws balanced!