A free neutron decays into a proton, an electron and
A.
a beta particle
B.
an alpha particle
C.
an antineutrino
D.
a neutrino
Answer :
an antineutrino
Solution :
Pauli suggested that after emission of $$\beta $$-particle (electron) a neutron is converted into a proton in a nucleus and in this reaction an electron and an antineutrino $$\left( {\overline v } \right)$$ will be formed. This reaction is represented as
\[\begin{array}{*{20}{c}}
{_0{n^1}}\\
{\left( {{\rm{Neutron}}} \right)}
\end{array} \to \begin{array}{*{20}{c}}
{_1{H^1}}\\
{\left( {{\rm{Proton}}} \right)}
\end{array} + \begin{array}{*{20}{c}}
{_{ - 1}{\beta ^0}}\\
{\left( {{\rm{Electron}}} \right)}
\end{array} + \begin{array}{*{20}{c}}
{\bar \upsilon }\\
{\left( {{\rm{Antineutrino}}} \right)}
\end{array}\]
Antineutrino is a particle whose mass is negligible and on which no charge is present. NOTE
After emission of $$\beta $$-particle, the total number of particles (mass-number) in a nucleus remains unchanged but no. of neutrons reduces by 1 making the no. of protons (i.e. charge-number) to increase by 1.
Releted MCQ Question on Modern Physics >> Atoms or Nuclear Fission and Fusion
In the nuclear fusion reaction
$$_1^2H + _1^3H \to _2^4He + n$$
given that the repulsive potential energy between the two
nuclei is $$ \sim 7.7 \times {10^{ - 14}}J,$$ the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s Constant $$k = 1.38 \times {10^{ - 23}}J/K$$ ]
The binding energy per nucleon of deuteron $$\left( {_1^2H} \right)$$ and helium nucleus $$\left( {_2^4He} \right)$$ is $$1.1\,MeV$$ and $$7\,MeV$$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is