QUESTION:
A free neutron can decay into a proton, an electron, and an anti-neutrino. Assume the anti-neutrino’s rest mass is zero, and the rest masses for proton and electron are 1.6726 \times 10^{ -27}\rm \ kg and 9.11\times10^{ -31}\rm \ kg respectively. Determine the total kinetic energy shared among the three particles when a neutron decays at rest.
SOLUTION:
Given:
- the rest masses of the proton is 1.6726\times 10^{ -27}\rm \ kg,
- the rest masses of the electron is 9.11\times 10^{ -31}\rm \ kg, and
the rest masses of the anti-neutrino is 0.
To find:
- the total kinetic energy shared among the three particles when a neutron decays at rest.
We know the values:
- the rest masses for neutron is 1.6749 \times 10^{ -27}\rm \ kg, and
- the unified atomic mass, \rm u = 1.66054\times 10^{ -27}\rm \ kg
The nuclear reaction of neutron-decay is
\rm n \rightarrow p + e^{-1} + \bar{\nu}
The mass defect of the above nuclear reaction is given by;
\begin{aligned} \Delta m &= \text{Mass}(\rm neutron) – \text{Mass} (\rm protron)- \text{Mass} (\rm electron) -Mass (\rm anti-neutrino) \\ &=(1.674929\times 10^{ -27}\rm \ kg) – (1.6726\times 10^{ -27} \ kg)-(9.11\times 10^{ -31} \ kg)-0\\ &= 0.001418\rm \ kg\\ \end{aligned}
Therefore, the total kinetic energy shared among the three particles when a neutron decays at rest is;
\begin{aligned} K.E._{( p + e^{-1} + \bar{\nu})} &= \Delta m c^2\\ &=(0.001418\rm \ kg) \times (3\times 10^8 \ m/s)^2\\ &= 1.2762\times 10^6\rm \ J \\ \end{aligned}
Similar Problems based on Mass-Energy Equivalence
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The isotope \rm ^{218}Po decays via \alpha-decay. The measured atomic mass of \rm ^{218}Po is 218.00897\rm \ u, and the atomic mass of the daughter nucleus is 213.99981\rm \ u. Find 1) the Name of the daughter nucleus, 2) the number of nucleons in the daughter nucleus, 3) the atomic number of the daughter nucleus, 4) the number of neutrons in the daughter nucleus, and 5) the kinetic energy of the \alpha-particle. (Ignore the recoil of the daughter nucleus.)
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