nuclear equations for alpha beta and gamma decay - The nucleus ^{23}_{10} Ne decays by \beta^{-} emission. Write down the \beta -decay equation and determine the maximum kinetic energy of the electrons emitted. Given that: m ( ^{23}_{10} Ne ) = 22.994466 u m (^{23}_{11} Na ) = 22.089770 u.

The nucleus ^{23}_{10} Ne decays by \beta^{-} emission. Write down the \beta -decay equation and determine the maximum kinetic energy of the electrons emitted. Given that: m ( ^{23}_{10} Ne ) = 22.994466 u m (^{23}_{11} Na ) = 22.089770 u.

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Anonym0us (Student, UG1 Grade)

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Positron emission - Wikipedia

Positron emission or beta plus decay is a subtype of radioactive decay called beta decay, in which a proton inside a radionuclide nucleus is converted into a ...

For more information, see Positron emission - Wikipedia

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How to tell what type of decay a radioactive element will undergo ...

Nuclear structure and decay data web site and app from the IAEA gives you detailed and color ... Radioactive decay is driven by the number of protons and neutrons in the nucleus. ... Virgil Fenn, studied Mathematics & Science at Public Libraries ... A carbon-14 atom decays by emitting a beta particle (an electron) and an ...

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Nuclei

nucleus of an α-particle of kinetic energy 5.5 MeV is about 4.0 × 10–14 m. ... 13.4 MASS-ENERGY AND NUCLEAR BINDING ENERGY ... Here the energy equivalent of mass m is related by the above equation ... Using this, express the mass defect ... In any radioactive sample, which undergoes α, β or γ-decay, it is found.

For more information, see Nuclei

Teja (last edited 1 month ago) (Student, UG2 Grade)

When a neutron is converted into a proton, the process is known as beta decay.

The Beta decay general equation:

_6^{14}C\ \ \rightarrow_{\ \ \ 7}^{\ \ 14}N\ +\ _{-1}^{\ \ 0}e\ +\ \overline{v}\ +Q

\text{ Q - value } = \text{ mass of } ^{14}_{6} C - \text{ mass of } ^{14}_{7} N

Neglecting the mass of electron (_{-1}^{0}e ) and antineutron \bar{v} .

Step 1: Writing the nuclear equation for beta decay

\beta - decay of ^{23}_{10} Ne

^{23}_{10} Ne \rightarrow ^{23}_{11} Na - {_{-1}^{0}e} + \bar{v} + Q

\text{ Q - value } = \text{ mass of } ^{23}_{10} Ne - \text{ mass of } ^{23}_{11} Na

Step 2: Determine the maximum kinetic energy(Q-value) of the electrons

Given that

Mass of ^{23}_{10} Ne = 22.994466 u

Mass of ^{23}_{11} Na = 22.089770 u

\text{ Q - value } = 22.994466 - 22.089770 u

\text{ Q - value } = 0.004696 c^2 u

\text{ Q - value } = 0.004696 c^2 * 931.5 MeV/c^2 \because 1 u = 931.5 MeV/c^2

\text{ Q - value } = 4.374 MeV

Hence, kinetic energy of electron (_{-1}^{0}e ) and antineutron \bar{v} = 4.374 MeV

If \bar{ v } = 0 , then maximum kinetic energy of electrons emitted = 4.374 MeV