What fraction of the 40 K that was on Earth when it formed \(4.5 \times {10^9}\,{\rm{year}}\) ago is left today?

A fraction is a portion of a whole or, more broadly, any number of equal parts. In everyday English, a fraction describes the number of parts of a specific size, such as one-half, eight-fifths, or three-quarters.

Consider the original figure of\(^{40}\;{\rm{K}}\)the current number is\({N_0}\), and the atom is\(N\).

As a result, we have according to the radioactive decay equation.

\(\begin{array}{c}N = {N_0}{e^{ - \lambda t}}\\\frac{N}{{{N_0}}} = {e^{ - \lambda t}}\end{array}\)

The decay constant,\(\lambda \), is defined as follows:

\(\lambda = \frac{{0.693}}{{{t_{1/2}}}}\)

The half-life of the radioactive element is\({t_{1/2}}\).

We now have\(^{40}\;{\rm{K}}\).

\({t_{1/2}} = 1.28 \times {10^9}\,{\rm{y}}\)

And\(t\)is the time

\(t = 4.5 \times {10^9}\,{\rm{y}}\)

As a result, the fraction is

\(\begin{array}{c}\frac{{{N_0}}}{N} = {e^{ - \frac{{0.693}}{{{L_{1/2}}}}t}}\\ = {e^{ - \frac{{0.693}}{{\left( {1.28 \times {{10}^9}\,y} \right)}}\left( {4.5 \times {{10}^9}\,{\rm{y}}} \right)}}\\ = 0.0874\end{array}\)

Therefore

Hence, just \(0.0874\) of the original \(^{40}\;{\rm{K}}\) is remaining today.