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Q17E

Expert-verifiedFound in: Page 518

Book edition
2nd Edition

Author(s)
Randy Harris

Pages
633 pages

ISBN
9780805303087

**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 $150{E}_{I}$.

Atomic number of nucleus, $A=220$ .

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

Binding energy of interior nucleon is ${E}_{I}$.

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

**Area of the nucleus is given as:**

** $\begin{array}{l}{a}_{nucleus}=4\pi {r}^{2}\\ {a}_{nucleus}=4\pi {\left({A}^{1/3}{R}_{o}\right)}^{2}\\ {a}_{nucleus}=4\pi {A}^{2/3}{R}_{o}^{2}\end{array}$**

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

(a)

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

Surface area of the nucleus is given as:

$\begin{array}{l}{a}_{\text{nucleus}}=4\pi {r}^{2}\\ {a}_{\text{nucleus}}=4\pi {\left({A}^{1/3}{R}_{o}\right)}^{2}\\ {a}_{\text{nucleus}}=4\pi {A}^{2/3}{{R}^{2}}_{o}\end{array}$

.

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

$\begin{array}{l}N=\frac{{a}_{\text{nucleus}}}{{a}_{nucleus}}\\ N=\frac{4{\pi}^{2/3}{R}_{0}^{2}}{\pi {R}_{o}^{2}}\end{array}$

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

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

(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 $\left(\frac{{E}_{l}}{2}\right)$ .

The total binding energy would be given as follows:

Binding energy $=75{E}_{I}+145\left(\frac{{E}_{I}}{2}\right)=147.5{E}_{I}$.

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