# When 2310Ne (mass = 22.9945 u) decays to 2311Na (mass = 22.9898 u), what is the maximum kinetic...

## Question:

When 2310Ne (mass = 22.9945 u) decays to 2311Na (mass = 22.9898 u), what is the maximum kinetic energy of the emitted electron?

## Kinetic Energy

The energy that is developed in a body due to the motion it undertakes while shifting from one point to another is referred to as kinetic energy. It is not dependent on the position of the body but is dependent on the movement. It is expressed in the unit of Joules in the SI unit of the system. Another unit to quantify its magnitude is MeV.

Given data

• The mass of 2310Ne is: {eq}{m_i} = 22.9945\;{\rm{u}} {/eq}.
• The mass of 2311Ne is: {eq}{m_f} = 22.9898\;{\rm{u}} {/eq}.

The expression for the mass defect is given by,

{eq}{M_D} = {m_i} - {m_f} {/eq}

Substitute the values in the above equation.

{eq}\begin{align*} {M_D} &= 22.9945\;{\rm{u}} - 22.9898\;{\rm{u}}\\ &= 0.0047\;{\rm{u}} \end{align*} {/eq}

It is known that {eq}1\;{\rm{u}} {/eq} equals approximately {eq}931.5\;{\rm{MeV}} {/eq} of energy.

The maximum kinetic energy of the electron that is emitted becomes.

{eq}\begin{align*} {E_K} &= {M_D} \times 931.5\;{\rm{MeV}}\\ &= \left( {0.0047} \right) \times 931.5\;{\rm{MeV}}\\ &= {\rm{4}}{\rm{.37805}}\;{\rm{MeV}} \end{align*} {/eq}

Thus, the maximum kinetic energy of the electron emitted is {eq}{\rm{4}}{\rm{.37805}}\;{\rm{MeV}} {/eq}.