Chapter 10, Problem 10C.3E

(a)

Interpretation Introduction

Interpretation:

The nuclear binding energy of nickel-62 nucleus has to be given.

Concept Introduction:

Nuclear binding energy and the mass defect:

Nuclear binding energy is the minimum energy that would be required to disassemble the nucleus of an atom into its component parts.  These component parts are neutrons and protons which are collectively called nucleons.  The mass of an atomic nucleus is less than the sum of the individual masses of the free constituent protons and neutrons according to Einstein’s equation ${\text{E=mc}}^{\text{2}}$.  The missing mass is known as the mass defect and represents the energy that was released when the nucleus was formed.

  $\Delta \text{m=m}\left(\text{nucleus}\right)\text{-}\sum \text{m}\left(\text{nucleons}\right)$

Answer to Problem 10C.3E

The binding energy of the carbon nucleus is $\text{8}{\text{.7425×10}}^{\text{-11}}\text{J}\text{.}$

Explanation of Solution

Given:

The mass of nickel-62 is $\text{61}\text{.928346u}\text{.}$

The mass of neutron is $\text{1}\text{.008664u}\text{.}$

The mass of proton is $\text{1}\text{.007276u}\text{.}$

The mass of an electron is $\text{0}\text{.00054858u}\text{.}$

The mass defect can be calculated as,

  $\Delta \text{m=m}\left(\text{nucleus}\right)\text{-}\sum \text{m}\left(\text{nucleons}\right)$

  $\begin{array}{l}\Delta \text{m=61}\text{.928346-}\left\{\left(\text{34×1}\text{.00864u}\right)\text{+}\left(\text{28×1}\text{.007276u}\right)\text{+}\left(\text{28×0}\text{.0005485u}\right)\right\}\\ \Delta \text{m=-0}\text{.5850×1}{\text{.6605×10}}^{\text{-27}}\text{kg}\\ \Delta \text{m=}\text{-}\text{9}{\text{.7139×10}}^{\text{-28}}\text{kg}\text{.}\end{array}$

The binding energy of the carbon nucleus is,

  $\begin{array}{l}\text{ΔE=}\left|\text{9}{\text{.7139×10}}^{\text{-28}}\right|\text{×}{\left({\text{3×10}}^{\text{8}}\right)}^{\text{2}}\\ \text{ΔE=8}{\text{.7425×10}}^{\text{-11}}\text{J}\end{array}$

The binding energy of the Nickel-62 nucleus is $\text{8}{\text{.7425×10}}^{\text{-11}}\text{J}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The binding energy of Plutonium-239 has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10C.3E

The binding energy of the plutonium-239 nucleus is $\text{2}{\text{.8973×10}}^{\text{-11}}\text{J}\text{.}$

Explanation of Solution

Given:

The mass of plutonium is $\text{239}\text{.0522u}\text{.}$

The mass of neutron is $\text{1}\text{.008664u}\text{.}$

The mass of proton is $\text{1}\text{.007276u}\text{.}$

The mass of an electron is $\text{0}\text{.00054858u}\text{.}$

The mass defect can be calculated as,

  $\Delta \text{m=m}\left(\text{nucleus}\right)\text{-}\sum \text{m}\left(\text{nucleons}\right)$

  $\begin{array}{l}\Delta \text{m=239}\text{.0522-}\left\{\left(\text{145×1}\text{.00864u}\right)\text{+}\left(\text{94×1}\text{.007276u}\right)\text{+}\left(\text{94×0}\text{.0005485u}\right)\right\}\\ \Delta \text{m=}\text{-1}\text{.9387×1}{\text{.6605×10}}^{\text{-27}}\text{kg}\\ \Delta \text{m=}\text{-}3.2192{\text{×10}}^{\text{-27}}\text{kg}\text{.}\end{array}$

The binding energy of the carbon nucleus is,

  $\begin{array}{l}\text{ΔE=}\left|\text{3}{\text{.2192×10}}^{\text{-27}}\right|\text{×}{\left({\text{3×10}}^{\text{8}}\right)}^{\text{2}}\\ \text{ΔE=2}{\text{.8973×10}}^{\text{-11}}\text{J}\end{array}$

The binding energy of the plutonium-239 nucleus is $\text{2}{\text{.8973×10}}^{\text{-11}}\text{J}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The binding energy of Deuterium has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10C.3E

The binding energy of the Deuterium nucleus is $\text{3}{\text{.55×10}}^{\text{-13}}\text{J}\text{.}$

Explanation of Solution

Given:

The mass of Deuterium is $\text{2}\text{.0141u}\text{.}$

The mass of neutron is $\text{1}\text{.008664u}\text{.}$

The mass of proton is $\text{1}\text{.007276u}\text{.}$

The mass of an electron is $\text{0}\text{.00054858u}\text{.}$

The mass defect can be calculated as,

  $\Delta \text{m=m}\left(\text{nucleus}\right)\text{-}\sum \text{m}\left(\text{nucleons}\right)$

  $\begin{array}{l}\text{Δm=2}\text{.0141-}\left\{\left(\text{1×1}\text{.00864u}\right)\text{+}\left(\text{1×1}\text{.007276u}\right)\text{+}\left(\text{1×0}\text{.0005485u}\right)\right\}\\ \text{Δm=}\text{-0}\text{.00238×1}{\text{.6605×10}}^{\text{-27}}\text{kg}\\ \text{Δm=}\text{-}\text{3}{\text{.951×10}}^{\text{-27}}\text{kg}\text{.}\end{array}$

The binding energy of the carbon nucleus is,

  $\begin{array}{l}\text{ΔE=}\left|\text{3}{\text{.951×10}}^{\text{-27}}\right|\text{×}{\left({\text{3×10}}^{\text{8}}\right)}^{\text{2}}\\ \text{ΔE=3}{\text{.55×10}}^{\text{-13}}\text{J}\end{array}$

The binding energy of the Deuterium nucleus is $\text{3}{\text{.55×10}}^{\text{-13}}\text{J}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The binding energy of tritium-3 has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10C.3E

The binding energy of the tritium nucleus is $\text{1}{\text{.359×10}}^{\text{-12}}\text{J}\text{.}$

Explanation of Solution

Given:

The mass of Tritium is $\text{3}\text{.01605u}\text{.}$

The mass of neutron is $\text{1}\text{.008664u}\text{.}$

The mass of proton is $\text{1}\text{.007276u}\text{.}$

The mass of an electron is $\text{0}\text{.00054858u}\text{.}$

The mass defect can be calculated as,

  $\Delta \text{m=m}\left(\text{nucleus}\right)\text{-}\sum \text{m}\left(\text{nucleons}\right)$

  $\begin{array}{l}\text{Δm=3}\text{.01605-}\left\{\left(\text{2×1}\text{.00864u}\right)\text{+}\left(\text{1×1}\text{.007276u}\right)\text{+}\left(\text{1×0}\text{.0005485u}\right)\right\}\\ \text{Δm=}\text{-0}\text{.009094×1}{\text{.6605×10}}^{\text{-27}}\text{kg}\\ \text{Δm=}\text{-}1.5100{\text{×10}}^{\text{-29}}\text{kg}\text{.}\end{array}$

The binding energy of the carbon nucleus is,

  $\begin{array}{l}\text{ΔE=}\left|\text{1}{\text{.5100×10}}^{\text{-29}}\right|\text{×}{\left({\text{9×10}}^{\text{8}}\right)}^{\text{2}}\\ \text{ΔE=1}{\text{.359×10}}^{\text{-12}}\text{J}\end{array}$

The binding energy of the Deuterium nucleus is $\text{1}{\text{.359×10}}^{\text{-12}}\text{J}\text{.}$

(e)

Interpretation Introduction

Interpretation:

The most stable nuclide has to be given.

Concept Introduction:

The binding energy is directly proportional to the stability of the nuclide.  Thus, greater the binding energy greater will be the stability of the nucleus.

Answer to Problem 10C.3E

Nickel -62 is the most stable nucleus.

Explanation of Solution

Since, the binding energy of the nickel-62 is greater it is the most stable nucleus.