Question

A \(_5^8\;{\rm{B}}\) atom (mass \( = 8.0246{\rm{amu}}\) ) decays into a\(_4^8\;{\rm{B}}\)atom (mass \(\left. { = 8.0053{\rm{amu}}} \right)\) by loss of a \({\beta ^ + }\)particle (mass \( = \)\(0.00055{\rm{amu}}\) ) or by electron capture. How much energy (in millions of electron volts) is produced by this reaction?

Short answer

\(17.37{\rm{MeV}}\) energy (in millions of electron volts) is produced by this reaction.

Step-by-step solution

Definition of mass

A physical body's mass is the amount of matter it contains. It's also a measure of the body's inertia or resistance to acceleration (velocity change) when a net force is applied.

Convert \({\bf{amu}}\) into \({\rm{kg}}\)

Let us solve the given problem.

We have this reaction:

\(_5^8Be \to _4^8B + _{ + 1}^0e\)

We can get the energy from this formula:

\(\begin{array}{c}E = m \cdot {c^2}\\m = m({\rm{ reactants }}) - m({\rm{ products }})\\\,m = 8.0246{\rm{amu}} - (8.0053{\rm{amu}} + 0.00055{\rm{amu}})\,\\m = 0.01875{\rm{amu}}\end{array}\)

We need to convert amu into\({\rm{kg}}\):

\(\begin{array}{c}1{\rm{amu}} = 1.6605 \cdot {10^{ - 27}}\;{\rm{kg}}\\m = 0.01875{\rm{amu}} \cdot 1.6605 \cdot {10^{ - 27}}\;{\rm{kg}} \cdot {\rm{am}}{{\rm{u}}^{ - 1}}\\m = 3.113 \cdot {10^{ - 29}}\;{\rm{kg}}\end{array}\)

Therefore, \(m = 3.113 \cdot {10^{ - 29}}\;{\rm{kg}}\).

Step 2: Transfer \(J\) into \({\rm{MeV}}\)

Consider the given problem.

Now we can get the energy:

\(\begin{array}{l}E = m \cdot {c^2}\\E = 3.113 \cdot {10^{ - 29}}\;{\rm{kg}} \cdot 2.99 \cdot {10^8}\;{\rm{m}}{{\rm{s}}^{ - {1^2}}}\\E = 2.783 \cdot {10^{ - 12}}\;{\rm{J}}\end{array}\)

But we need to transfer\(J\)into\({\rm{MeV}}\):

Now we need to convert the energy from\(J\)to\(MeV\):

\(1{\rm{MeV}} = 1.602 \cdot {10^{ - 13}}\;{\rm{J}}\)

\(E = 2.783 \cdot {10^{ - 12}}\;{\rm{J}} \cdot \frac{{1{\rm{MeV}}}}{{1.602 \cdot {{10}^{ - 13}}\;{\rm{J}}}}\)

Therefore,

The required energy is\(17.37{\rm{MeV}}\).