Question

During a $\beta $ positive decay experiment it is observed that kinetic energy of $\beta $ positive particle is $50\% $ of $Q$ value of the reaction, then select the correct alternative.
A. Almost the remaining $50\% $ energy must be in the form of kinetic energy of the daughter nucleus.
B. Almost the remaining $50\% $ energy must be lost in the form of heat.
C. Almost the remaining $50\% $ energy must be taken away by neutrinos.
D. Almost the remaining $50\% $ energy must be taken away by antineutrino.

Answer 1

Hint: In order to solve this problem let us first get some idea about radioactive. Radioactivity is the property of certain forms of matter to spontaneously emit energy and subatomic particles. It is essentially a property of individual atomic nuclei.

Complete step by step solution:
Positron emission, also known as positive beta decay \[\left( {\beta + - decay} \right),\]occurs when a proton in the parent nucleus decays into a neutron that remains in the daughter nucleus, and the nucleus emits a neutrino and a positron, a positive particle with the same mass as an ordinary electron but the opposite charge. As a result, positive beta decay produces a daughter nucleus with an atomic number one lower than its parent but the same mass number. Irene and Frederic Joliot-Curie discovered positron emission for the first time in \[1934\].
\[_Z{X^A}{ \to _{Z - 1}}{Y^A}{ + _{ - 1}}{e^0} + v + Q\]
Almost majority of the energy generated by \[Q\] value is dispersed among the participants $_{ + 1}{e^0}\& \,v$
So, option C is correct.

Note:
Let us know some more important points regarding beta decay .Beta decay is a sluggish process when compared to other types of radioactivity, such as gamma or alpha decay. Beta decay half-lives are never less than a few milliseconds.