Q80PE. - College Physics (Urone)

Question

The relatively scarce naturally occurring calcium isotope \(^{48}{\rm{Ca}}\) has a half-life of about \(2 \times {10^{16}}{\rm{y}}\). (a) A small sample of this isotope is labeled as having an activity of \(1.0\)Ci. What is the mass of the \(^{48}{\rm{Ca}}\)in the sample?

(b) What is unreasonable about this result? 

(c) What assumption is responsible?

Step-by-Step Solution

Answer

a) The mass of the \(^{48}{\rm{Ca}}\) in the sample is \(3.0 \times {10^9}\;{\rm{kg}}\).

b) The unreasonable about this result is it has a very high value.

c) The idea that half-life is a finite amount of time is \(2 \times {10^{16}}{\rm{y}}\) which is high.

Definition of isotope

Isotopes are atoms with variable numbers of neutrons but the same number of protons and electrons in the same element.

The number of neutrons in the various isotopes of an element differs, implying that the isotopes have varying masses.

The mass of the \(^{48}{\rm{Ca}}\) in the sample

(a)

Let us solve the given problem.

The relation between activity, half-life and the number of atoms is given by:

\(R = \frac{{0.693N}}{{{t_{1/2}}}} \to N = \frac{{R{t_{1/2}}}}{{0.693}}\)

Where \({t_{1/2}} = \) half-life , \({\rm{R}} = \)activity and \(N = \)number of atoms.

Also we know that \(1\,{\rm{Ci}} = 3.70 \times {10^{10}}\;\,{\rm{Bq}}\) and,

 \(\begin{aligned}{t_{1/2}} = \left( {2 \times {{10}^{16}}{\rm{y}}} \right)\left( {3.17 \times {{10}^7}\;{\rm{s}}} \right)\\ = 6.34 \times {10^{23}}\;{\rm{s}}\end{aligned}\)

Now plug in the value of \({\rm{R}}\)and \({t_{1/2}}\)in the above equation and solve for the value of 

\({\rm{N}}\):

\(\begin{aligned}N = \frac{{\left( {3.70 \times {{10}^{10}}\;{\rm{Bq}}} \right)\left( {6.34 \times {{10}^{23}}\;{\rm{s}}} \right)}}{{0.693}}\\ = 3.38 \times {10^{34}}\\N = 3.38 \times {10^{34}}\end{aligned}\)

Now we have to convert number of atoms into mass by the given formula:

\(m = N\left( {\frac{{1\;{\rm{mol}}}}{{6.02 \times {{10}^2}3}}} \right)\left( {\frac{M}{{1\;{\rm{mol}}}}} \right)\)

Now plugin the value of \(N\) and \(48{\rm{G}}/{\rm{mol}}\) for \(M\) and solve for the value of \(m\)

\(\begin{aligned}m = \left( {3.38 \times {{10}^{34}}} \right)\left( {\frac{{1\;{\rm{mol}}}}{{6.02 \times {{10}^{23}}}}} \right)\left( {\frac{{48}}{{1\;{\rm{mol}}}}} \right)\\ = 3.0 \times {10^{12}}\;{\rm{g}}\\ = 3.0 \times {10^9}\;{\rm{kg}}\\m = 3.0 \times {10^9}\;{\rm{kg}}\end{aligned}\)

Therefore, the mass is \(3.0 \times {10^9}\;{\rm{kg}}\).

Unreasonable about result in part a)

b)

The value of mass is very high.

Assumption which is responsible

c)

The assumption that half-life is \(2 \times {10^{16}}{\rm{y}}\) high.