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, $n\to p+e_{-1}^0+\bar{\nu }$
$m_n{c}^2=m_p{c}^2+m_e{c}^2+m_{an}{c}^2$
${\text{m}}_{an}{\text{c}}^2=(m_n-m_p-m_e)c^2$
$\text{ = (1}\text{.00866u - 1}\text{.00727 - 0}\text{.00055u)}{c}^2$
$\text{ = }0.00084u{c}^2$
$\text{ = 0}\text{.00084}\times \text{931}\text{.5MeV}$
$m_{an}{c}^2=0.78\times {10}^6\text{eV}$
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.
$m_n{c}^2=m_p{c}^2+m_e{c}^2+m_{an}{c}^2$
${\text{m}}_{an}{\text{c}}^2=(m_n-m_p-m_e)c^2$
$\text{ = (1}\text{.00866u - 1}\text{.00727 - 0}\text{.00055u)}{c}^2$
$\text{ = }0.00084u{c}^2$
$\text{ = 0}\text{.00084}\times \text{931}\text{.5MeV}$
$m_{an}{c}^2=0.78\times {10}^6\text{eV}$