Question

What is the maximum energy of the anti- neutrino?
A. zero
B. Much less than $0.8 \times {10^6}{\text{eV}}$
C. Nearly $0.8 \times {10^6}{\text{eV}}$
D. much larger than $0.8 \times {10^6}{\text{eV}}$

Answer 1

Hint: Using Einstein’s mass energy equation,$E = m{c^2}$solve the given problem.
Formula used: ${m_n}{c^2} = {m_p}{c^2} + {m_e}{c^2} + {m_{an}}{c^2}$

Complete step-by-step answer:
We can consider the Einstein’s mass- energy relation,
The beta decay is a negative process.
The negative $\beta $-decay, $n \to p + e_{ - 1}^0 + \overline \nu $
${m_n}{c^2} = {m_p}{c^2} + {m_e}{c^2} + {m_{an}}{c^2}$
${{\text{m}}_{an}}{{\text{c}}^2} = ({m_n} - {m_p} - {m_e}){c^2}$
${\text{ = (1}}{\text{.00866u - 1}}{\text{.00727 - 0}}{\text{.00055u)}}{c^2}$
${\text{ = }}0.00084u{c^2}$
${\text{ = 0}}{\text{.00084}} \times {\text{931}}{\text{.5MeV}}$
${m_{an}}{c^2} = 0.78 \times {10^6}{\text{eV}}$
Which is less than $0.8 \times {10^6}{\text{eV}}$.

From the above equation we found the value for the maximum energy of the anti-neutrino.
Therefore, answer is (B) Much less than $0.8 \times {10^6}{\text{eV}}$.

Additional information:
There are three types of the decay process
Alpha decay
Beta decay
Gamma decay
The process of emission of an electron or positron from a radioactive nucleus is called the $\beta - {\text{decay}}$.
When a nucleus emits a $\beta - $particle (electron or positron), the mass number A of the nucleus does not change but its atomic number Z increases by 1. Therefore, $\beta - {\text{decay}}$can be represented as:
\[{\text{X}}_Z^A \to {\text{Y}}_{Z + 1}^A + e_{ - 1}^0 + Q\]
where Q represents the maximum kinetic energy with which the $\beta $- particles are emitted.$Q = \left( {{m_x} - {m_y}} \right){c^2}$

According to the above relation all -particles must be emitted with the same kinetic energy. But the experiments showed that only a few -particles are emitted with the maximum kinetic energy. A large number of - particles are emitted with a small value of the kinetic energy. Thus, there is an apparent break or violation of the law of conservation of mass-energy. To overcome this difficulty Pauli proposed a theory called the “neutrino hypothesis”.
According to this hypothesis, there must exist another particle called “neutrino” to account for the missing energy and momentum. Therefore, in a negative -decay, a neutron is transformed into a proton, electron, and an anti-neutrino. That is, Similarly, in a positive -decay, a proton is transferred into a neutron, positron, and a neutrino. That is,
Neutrinos are neutral particles with very small mass compared to the electron. They have only weak interactions with other particles. Therefore, they are very difficult to detect.

Note: By learning the atomic mass unit of all the fundamental particles the above problem can be solved.
${m_n}{c^2} = {m_p}{c^2} + {m_e}{c^2} + {m_{an}}{c^2}$
${{\text{m}}_{an}}{{\text{c}}^2} = ({m_n} - {m_p} - {m_e}){c^2}$
${\text{ = (1}}{\text{.00866u - 1}}{\text{.00727 - 0}}{\text{.00055u)}}{c^2}$
${\text{ = }}0.00084u{c^2}$
${\text{ = 0}}{\text{.00084}} \times {\text{931}}{\text{.5MeV}}$
${m_{an}}{c^2} = 0.78 \times {10^6}{\text{eV}}$