Book edition 2nd Edition

Author(s) Randy Harris

Pages 633 pages

ISBN 9780805303087

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Short Answer

a) For a nucleus of $A = 220$ , estimate very roughly how many nucleons would be at the surface.
b) If the binding energy of an interior nucleon due to the internucleon attraction were $E_{1}$ and if all nucleon were are tightly bound, the total binding energy would be $A E_{1}$ . Taking into account the different binding of surface nucleons, roughly what would be the total binding energy?

a) The number of nucleons at the surface of the nucleus is roughly $145$ .

b) The total binding energy of nucleus is approximate $150 E_{I}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

Atomic number of nucleus, $A = 220$ .

Radius of nucleons, $R_{0}$ is $1.2 \times 10^{- 15} m$ .

Binding energy of interior nucleon is $E_{I}$ .

Binding energy of surface nucleon is $E_{I} / 2$ .

Step 2: Formula of Area of nucleus

Area of the nucleus is given as:

$\begin{array}{l} a_{n u c l e u s} = 4 π r^{2} \\ a_{n u c l e u s} = 4 π {(A^{1 / 3} R_{o})}^{2} \\ a_{n u c l e u s} = 4 π A^{2 / 3} R_{o}^{2} \end{array}$

Where, $r$ is the radius of large sphere, $R o$ is the radius of nucleon.

Step 3: Calculation for the number of nucleon

(a)

The area contribution of each nucleon present on the boundary surface is $π R_{o}^{2}$ where, $R_{o} = 1.2 \times 10^{- 15} m$ .

Surface area of the nucleus is given as:

$\begin{array}{l} a_{nucleus} = 4 π r^{2} \\ a_{nucleus} = 4 π {(A^{1 / 3} R_{o})}^{2} \\ a_{nucleus} = 4 π A^{2 / 3} {R^{2}}_{o} \end{array}$

The number of nucleon present on the surface as shown below.

$\begin{array}{l} N = \frac{a_{nucleus}}{a_{n u c l e u s}} \\ N = \frac{4 π^{2 / 3} R_{0}^{2}}{π R_{o}^{2}} \end{array}$

Substitute the values of area of nucleus in the above equation.

$\begin{array}{l} N = 4 A^{2 / 3} \\ N = 4 {(220)}^{2 / 3} \\ N = 145.77 \end{array}$

Step 4: Calculation for the binding energy of nucleon

(b)

The total number of nucleons in the interior of the nucleus is $220 - 145 = 75$ .

Binding energy of each interior nucleon is $E_{I}$ and each surface nucleon contributes about half as much binding energy, that is $(\frac{E_{l}}{2})$ .

The total binding energy would be given as follows:

Binding energy $= 75 E_{I} + 145 (\frac{E_{I}}{2}) = 147.5 E_{I}$ .

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Modern Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Formula of Area of nucleus

Step 3: Calculation for the number of nucleon

Step 4: Calculation for the binding energy of nucleon

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Modern Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Formula of Area of nucleus

Step 3: Calculation for the number of nucleon

Step 4: Calculation for the binding energy of nucleon

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