Question

a. Calculate the mass defect in Fe-56 if the mass of
an Fe-56 nucleus is 55.921 amu.  The mass of a
proton is 1.00728 amu and the mass of a neutron
is 1.008665 amu.

 

b. Determine the binding energy of an O-16
nucleus.  The O-16 nucleus has a mass of 15.9905
amu.  A proton has a mass of 1.00728 amu, a neutron has a mass of 1.008665 amu, and 1 amu is equivalent to 931 MeV of energy.

 

c. What percentage of a radioactive substance
remains after 7.00 half-lives have elapsed?

 

d. Carbon-11 is used in medical imaging. The half-
life of this radioisotope is 20.4 min. What
percentage of a sample remains after 60.0 min?

 

 

Answer 1

Concepts and reason

The concepts used to solve the given questions are based on the mass defect, binding energy and percentage of the remaining amount.

Firstly, mass defect in ${\rm{Fe}} - 56$ is calculated. Secondly, the binding energy of ${\rm{O}} - 16$ is determined and then the percentage of remaining amount is calculated using the given half-lives of the respective sample.

Fundamentals

Mass defect:

It is the difference between the mass of an isotope and its mass number. The unit of mass defect is (amu).

$1{\rm{ amu}} = 931{\rm{ MeV}}$

The formula of the mass defect $\left( {\Delta m} \right)$ as shown below.

$\Delta m = \left[ {Z\left( {{m_{\rm{p}}}} \right) + \left( {A - Z} \right){m_{\rm{n}}}} \right] - {m_{{\rm{atom}}}}$ …… (1)

Here, $Z$ is the atomic number, $A$ is the mass number, ${m_{\rm{p}}}$ is the mass of a proton, ${m_{\rm{n}}}$ is the mass of the neutron.

Binding energy:

This is energy which binds the nucleus together. It is equal to the mass defect.

${\rm{B}}{\rm{.E}} = \Delta m$ …… (2)

The number of half-lives can be calculated as follows.

$n = \frac{{{\rm{Time elapsed}}}}{{{\rm{half life}}}}$ …… (3)

Percentage of the remaining amount can be calculated as follows.

$\left( \% \right) = \frac{1}{{{2^{\rm{n}}}}} \times 100$ …… (4)

Here, $n$ is the number of half-lives.

(12.a)

To calculate the mass defect in ${\rm{Fe}} - 56$ , substitute the value of $Z$ as $26$ , $A$ as $56$ , ${m_{\rm{p}}}$ as $1.00728{\rm{ amu}}$ , ${m_{\rm{n}}}$ as $1.008665{\rm{ amu}}$ and ${m_{{\rm{atom}}}}$ as $55.921{\rm{ amu}}$ in the equation (1).

$\begin{array}{l}\\\Delta m = \left[ {26\left( {1.00728{\rm{ amu}}} \right) + \left( {56 - 26} \right)1.008665{\rm{ amu}}} \right] - 55.921{\rm{ amu}}\\\\\Delta m = \left[ {26.18928 + 30.25995{\rm{ }}} \right] - 55.921{\rm{ amu}}\\\\\Delta m = \left( {56.44923 - 55.921} \right){\rm{ amu}}\\\\\Delta m = 0.528{\rm{ amu}}\\\end{array}$

(12.b)

To calculate the mass defect in ${\rm{O}} - 16$ , substitute the value of $Z$ as $8$ , $A$ as $16$ , ${m_{\rm{p}}}$ as $1.00728{\rm{ amu}}$ , ${m_{\rm{n}}}$ as $1.008665{\rm{ amu}}$ and ${m_{{\rm{atom}}}}$ as $15.9905{\rm{ amu}}$ in the equation (1).

$\begin{array}{l}\\\Delta m = \left[ {8\left( {1.00728{\rm{ amu}}} \right) + \left( {16 - 8} \right)1.008665{\rm{ amu}}} \right] - 15.9905{\rm{ amu}}\\\\\Delta m = \left[ {8.05824 + 8.06932{\rm{ }}} \right] - 15.9905{\rm{ amu}}\\\\\Delta m = \left( {16.12756 - 15.9905} \right){\rm{ amu}}\\\\\Delta m = 0.13706{\rm{ amu}}\\\end{array}$

The binding energy is calculated as follows.

$\begin{array}{l}\\B.E = \Delta m \times {\rm{931 MeV}}\\\\B.E = \left( {0.13706 \times {\rm{931}}} \right){\rm{ MeV}}\\\\B.E = 127.6{\rm{ MeV}}\\\end{array}$

(12.c)

To calculate the percentage of the remaining amount of radioactive substance substitute the value of $n$ as $7$ in the equation (4).

$\begin{array}{c}\\\left( \% \right) = \frac{1}{{{2^7}}} \times 100\\\\ = 0.0078 \times 100\\\\ = 0.78\% \\\end{array}$

(12.d)

To calculate the value of the number of half-lives, substitute the value of half-life as $20.4\min$ and time elapsed as $60.0\min$ in the equation (3).

$\begin{array}{l}\\n = \frac{{60.0\min }}{{{\rm{20}}{\rm{.4}}\min }}\\\\n = 2.941\\\end{array}$

To calculate the percentage of the remaining amount of radioactive substance substitute the value of $n$ as $2.941$ in the equation (4).

$\begin{array}{c}\\\left( \% \right) = \frac{1}{{{2^{2.941}}}} \times 100\\\\ = 0.1302 \times 100\\\\ = 13.02\% \\\end{array}$

Ans: Part 12.a

The mass defect in ${\bf{Fe}} - {\bf{56}}$ is amu.

Part 12.b

The binding energy in ${\bf{O}} - {\bf{16}}$ is ${\bf{127}}{\bf{.6 MeV}}$ .

Part 12.c

The percentage of the remaining amount of radioactive substance is ${\bf{0}}{\bf{.78\% }}$ .

Part 12.d

The percentage of the remaining amount of radioisotope is ${\bf{13}}{\bf{.02\% }}$ .