Question

The β-decay process, discovered around 1900, is basically the decay of a neutron (n) in the laboratory. A proton (p) and an electron (e-) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three body decay process, i.e. $n \to p + {e^ - } + {\bar v_e}$ , around 1930, Pauli explained the observed electron energy spec trump. Assuming the anti-neutrino$\left( {{{\bar v}_e}} \right)$ to be massless and possessing negligible energy, and the neutron to be at rest, m0mentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of an electron is $0.8 \times {10^{ - 6}}eV$. The kinetic energy carried by the proton is only the recoil energy.
What is the maximum energy of the anti-neutrino?
A. Zero
B. Much less than $0.8 \times {10^6}eV$
C. Nearly $0.8 \times {10^6}eV$
D. Much larger than $0.8 \times {10^6}eV$

Answer 1

We solve this problem by using Einstein’s mass-energy relation (also given by special relativity theory). It states that mass is concentrated energy. The nuclear reaction releases more energy than the chemical reaction because of mass changes.

Complete step by step answer:
Each neutrino particle has an antiparticle called an antineutrino, which has no electric charge and half-integral spin. Antineutrino is produced in β- decay together with β particles, where the neutrino decays into proton, neutron, and antineutrino in the above theory.
Here the beta decay is a negative process.
$\therefore$ $n \to p + {e^ - } + {\bar v_e}$
Now, by using mass energy relation i.e. $E = m{c^2}$ for neutrino, proton, electron and antineutrino and we will get the result as,
$\Rightarrow {m_n}{c^2} = {m_p}{c^2} + {m_e}{c^2} + {m_{an}}{c^2}$
(Where mn is the mass of neutrino, mp mass of the proton, me mass of the electron and man is the mass of antineutrino, c is the speed of light.)
${m_{an}}{c^2} = \left( {{m_n} - {m_p} - {m_e}} \right){c^2}$
We can now substitute 1.00866u for 1.00727u for proton, and 0.00055u for electron and we will get,
$\Rightarrow {m_{an}}{c^2} = (1.00866u{\rm{ }} - 1.00727u{\rm{ }} - {\rm{ }}0.00055u){\rm{ }}{c^2}$
$= 0.00084u{c^2}$
Since the product of 1 amu and ${c^2}$ is equal to 931 MeV where c is the speed of light which is equal to ${c^2} = 3 \times {10^8}\;m/s$
Hence we can now simplify the above equation as,
${m_{an}}{c^2} = 0.00084u{c^2}$
${m_{an}}{c^2} = 0.00084 \times 931.5MeV$
${m_{an}}{c^2} = 0.78 \times {10^6}eV\gg {\rm{ }}0.8 \times {10^6}eV$

Therefore the maximum energy of antineutrino is nearly equal to ${\rm{ 0}}{\rm{.8}} \times {\rm{1}}{{\rm{0}}^6}{\rm{eV}}$ and the correct option is (B).

Additional information:
There are three types of decay processes exist i.e. Alpha decay, Beta decay, and Gamma decay. The process of emission of electrons or positrons of the radioactive nucleus called Beta decay. In beta decay, a large number of particles are emitted with the small value of kinetic energy and few particles have only the same kinetic energy. Thus, there is an apparent break of the law of mass-energy conservation. To overcome this problem Pauli’s proposed a theory called (neutrino –hypothesis).

Note:
In negative beta decay, the neutrino is transformed into an electron, proton, and an antineutrino. That is, neutrinos are neutral particles with very small mass compared to the electron. They have only weak interactions and it is very difficult to detect.