JEE Challenger
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PhyMedAdv 2016 • Paper 1

Determine Maximum Kinetic Energy of Beta Particle in Nuclear Decay

The isotope 512B{}_5^{12}\text{B} with atomic mass 12.014 u12.014\text{ u} undergoes β\beta-decay forming 612C{}_6^{12}\text{C}. An excited nuclear state 612C∗{}_6^{12}\text{C}^* exists at 4.041 MeV4.041\text{ MeV} above the ground state of 612C{}_6^{12}\text{C}. If 512B{}_5^{12}\text{B} decays into this excited state 612C∗{}_6^{12}\text{C}^*, calculate the maximum kinetic energy of the emitted β\beta-particle in units of MeV\text{MeV}. (Given: 1 u=931.5 MeV/c21\text{ u} = 931.5\text{ MeV}/c^2, where cc represents the speed of light in vacuum).

Official Numerical Answer8 to 9

Topics & Concepts

NucleiRadioactivityNuclear Reactions

Step-by-Step Solution

To determine the maximum kinetic energy of the emitted β\beta-particle, we analyze the nuclear β−\beta^--decay process:

512B→612C∗+e−+νˉe{}_5^{12}\text{B} \rightarrow {}_6^{12}\text{C}^* + e^- + \bar{\nu}_e

Step 1: Mass Defect and Total Q-Value for Ground State Decay

The atomic mass of carbon-12 (612C{}_6^{12}\text{C}) in its ground state is defined as exactly: M(612C)=12.000 uM({}_6^{12}\text{C}) = 12.000\text{ u}

The given atomic mass of boron-12 (512B{}_5^{12}\text{B}) is: M(512B)=12.014 uM({}_5^{12}\text{B}) = 12.014\text{ u}

In β−\beta^--decay, the difference in atomic masses automatically accounts for the mass of the emitted electron. Thus, the mass difference Δm\Delta m for decay to the ground state is: Δm=M(512B)−M(612C)=12.014 u−12.000 u=0.014 u\Delta m = M({}_5^{12}\text{B}) - M({}_6^{12}\text{C}) = 12.014\text{ u} - 12.000\text{ u} = 0.014\text{ u}

Using the energy equivalent 1 u=931.5 MeV/c21\text{ u} = 931.5\text{ MeV}/c^2, the total energy released (QQ-value) for decay to the ground state is: Q=Δm×931.5 MeV=0.014×931.5 MeV=13.041 MeVQ = \Delta m \times 931.5\text{ MeV} = 0.014 \times 931.5\text{ MeV} = 13.041\text{ MeV}

Step 2: Q-Value for Decay to the Excited State

The excited state 612C∗{}_6^{12}\text{C}^* lies at an energy level E∗=4.041 MeVE^* = 4.041\text{ MeV} above the ground state. Therefore, the available energy (Q∗Q^*) for the decay to this excited state is: Q∗=Q−E∗=13.041 MeV−4.041 MeV=9.000 MeVQ^* = Q - E^* = 13.041\text{ MeV} - 4.041\text{ MeV} = 9.000\text{ MeV}

Step 3: Maximum Kinetic Energy of the β\beta-Particle

The maximum kinetic energy of the emitted β\beta-particle occurs when the antineutrino carries away negligible kinetic energy. Neglecting the tiny recoil kinetic energy of the heavy daughter nucleus 612C∗{}_6^{12}\text{C}^*: Kβ,max=Q∗=9 MeVK_{\beta, \text{max}} = Q^* = 9\text{ MeV}

(Note: If one mistakenly accounts for electron mass subtraction as in β+\beta^+ decay, i.e., Q∗−2mec2≈9−1.022=7.978 MeV≈8 MeVQ^* - 2m_e c^2 \approx 9 - 1.022 = 7.978\text{ MeV} \approx 8\text{ MeV}, the rounded integer value yields 88. Hence, both 8 and 9 are acceptable depending on interpretation.)

Final Answer: The maximum kinetic energy of the β\beta-particle is 9 (or 8).

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