Question

In: Chemistry

a. Calculate the mass defect in Fe-56 if the mass ofan Fe-56 nucleus is 55.921 amu.  The...

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?

 

 

Solutions

Expert Solution

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 Fe56{\rm{Fe}} - 56 is calculated. Secondly, the binding energy of O16{\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).

1amu=931MeV1{\rm{ amu}} = 931{\rm{ MeV}}

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

Δm=[Z(mp)+(AZ)mn]matom\Delta m = \left[ {Z\left( {{m_{\rm{p}}}} \right) + \left( {A - Z} \right){m_{\rm{n}}}} \right] - {m_{{\rm{atom}}}} …… (1)

Here, ZZ is the atomic number, AA is the mass number, mp{m_{\rm{p}}} is the mass of a proton, mn{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.

B.E=Δm{\rm{B}}{\rm{.E}} = \Delta m …… (2)

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

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

Percentage of the remaining amount can be calculated as follows.

(%)=12n×100\left( \% \right) = \frac{1}{{{2^{\rm{n}}}}} \times 100 …… (4)

Here, nn is the number of half-lives.

(12.a)

To calculate the mass defect in Fe56{\rm{Fe}} - 56 , substitute the value of ZZ as 2626 , AA as 5656 , mp{m_{\rm{p}}} as 1.00728amu1.00728{\rm{ amu}} , mn{m_{\rm{n}}} as 1.008665amu1.008665{\rm{ amu}} and matom{m_{{\rm{atom}}}} as 55.921amu55.921{\rm{ amu}} in the equation (1).

Δm=[26(1.00728amu)+(5626)1.008665amu]55.921amuΔm=[26.18928+30.25995]55.921amuΔm=(56.4492355.921)amuΔm=0.528amu\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 O16{\rm{O}} - 16 , substitute the value of ZZ as 88 , AA as 1616 , mp{m_{\rm{p}}} as 1.00728amu1.00728{\rm{ amu}} , mn{m_{\rm{n}}} as 1.008665amu1.008665{\rm{ amu}} and matom{m_{{\rm{atom}}}} as 15.9905amu15.9905{\rm{ amu}} in the equation (1).

Δm=[8(1.00728amu)+(168)1.008665amu]15.9905amuΔm=[8.05824+8.06932]15.9905amuΔm=(16.1275615.9905)amuΔm=0.13706amu\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.

B.E=Δm×931MeVB.E=(0.13706×931)MeVB.E=127.6MeV\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 nn as 77 in the equation (4).

(%)=127×100=0.0078×100=0.78%\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 20.4min20.4\min and time elapsed as 60.0min60.0\min in the equation (3).

n=60.0min20.4minn=2.941\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 nn as 2.9412.941 in the equation (4).

(%)=122.941×100=0.1302×100=13.02%\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 Fe56{\bf{Fe}} - {\bf{56}} is amu.

Part 12.b

The binding energy in O16{\bf{O}} - {\bf{16}} is 127.6MeV{\bf{127}}{\bf{.6 MeV}} .

Part 12.c

The percentage of the remaining amount of radioactive substance is 0.78%{\bf{0}}{\bf{.78\% }} .

Part 12.d

The percentage of the remaining amount of radioisotope is 13.02%{\bf{13}}{\bf{.02\% }} .


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