Q31.5-52PE - College Physics (Urone)

Question

\({}^{{\rm{50}}}{\rm{V}}\) has one of the longest known radioactive half-lives. In a difficult experiment, a researcher found that the activity of \(1.00\,{\rm{kg}}\) of \({}^{{\rm{50}}}{\rm{V}}\)is \(1.75\,{\rm{Bq}}\). What is the half-life in years?

Step-by-Step Solution

Answer

The half-life is\(1.48 \times {10^{17}}\,{\rm{yr}}\).

Define radioactivity

Radioactivity is a phenomenon in which a few substances spontaneously release energy and subatomic particles. The nuclear instability of an atom causes radioactivity.

Evaluating the half life

The activity is performed by,

\(\begin{aligned}{c}R = \frac{{0.693N}}{{{t_{1/2}}}}\\{t_{1/2}} = \frac{{0.693N}}{R}\end{aligned}\)

 

\(R = 1.75\,{\rm{Bq}}\) is the given activity of the \({}^{{\rm{50}}}{\rm{V}}\)sample.

The provided sample has a mass of \(m = 1.00\,{\rm{kg}} = 1.00 \times {10^3}\,{\rm{g}}\).

\(M = 50.94\,{{\rm{g}} \mathord{\left/  {\vphantom {{\rm{g}} {{\rm{mol}}}}} \right.} {{\rm{mol}}}}\) is the molar mass of the element \({}^{{\rm{50}}}{\rm{V}}\). As a result, the total number of atoms in the sample is,

\(N = \frac{m}{M}{N_A}\)

 

Substitute all the value in the above equation

\(\begin{aligned}{\underline{\phantom{xx}}}N = \frac{{(1.00 \times {{10}^3}\,{\rm{g}})}}{{(50.94\,{{\rm{g}} \mathord{\left/ {\vphantom {{\rm{g}} {{\rm{mol}}}}} \right. }{{\rm{mol}}}})}}\left( {6.02 \times {{10}^{23}}\,{{{\rm{atoms}}} \mathord{\left/ {\vphantom {{{\rm{atoms}}} {{\rm{mol}}}}}\right.} {{\rm{mol}}}}} \right)\\N = 1.18178 \times {10^{25}}\,{\rm{atoms}}\end{aligned}\)

As a result, the half-life is equal to,

\(\begin{aligned}{\underline{\phantom{xx}}}{t_{1/2}} = \frac{{0.693(1.18178 \times {{10}^{25}}\,{\rm{atoms}})}}{{\left( {1.75\,{\rm{Bq}}} \right)}}\\ = 4.67985 \times {10^{24}}\,{\rm{s}}\\ = 1.48 \times {10^{17}}\,{\rm{y}}\end{aligned}\)

 

Therefore, the half-life is \(1.48 \times {10^{17}}\,{\rm{yr}}\).