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Q26E

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### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

# Nuclei of the same mass number but different Z are known as isobars. Oxygen-15 and nitrogen- 15 are isobars. (a) In which of the factors considered in nuclear binding (represented by terms in the semi empirical binding energy formula) do these two isobars differ?(b) Which of the isobars should be more tightly bound?(c) k your conclusion in part (b) supported by the decay mode information of Appendix 1? Explain.(d) Calculate the binding energies of oxygen-15 and nitrogen-15. By how much do they differ?(e) Repeat part (d) but use the semi empirical binding energy formula rather than the known atomic masses.

a) In the semi empirical binding energy formula, oxygen-15 and nitrogen-15 . The coulomb term and the last term, and asymmetry term are different.

b) The binding energy of nitrogen-15 is more and it is more tightly bound.

c) Yes, half-life decay mode support the nitrogen's nucleon's tightly bound statement

d) The binding energy of oxygen-15 is $111.96\text{MeV}$. The binding energy of nitrogen-15 is $115.49\text{MeV}$.

The difference is $3.53\text{MeV}$.

$\text{(e)}\Delta BE=4.39\text{MeV}$

See the step by step solution

## Step 1: Given data

Two Isobars, Oxygen- 15 and nitrogen- 15.

For oxygen-15 :

Atomic mass number, $A=15$.

Atomic number, Z=8.

Number of neutron, $N=A-Z=15-8=7$.

For nitrogen-15 :

Mass number, $A=15$.

Atomic number, $Z=7$.

Number of neutron, $N=A-Z=15-7=8$.

## Step 2: Formula of Semi empirical binding Energy

Semi empirical binding Energy is given by the expression,

${\mathbit{B}}{\mathbit{E}}{\mathbf{=}}{{\mathbit{c}}}_{{\mathbf{1}}}{\mathbit{A}}{\mathbf{-}}{{\mathbit{c}}}_{{\mathbf{2}}}{{\mathbit{A}}}^{\mathbf{2}\mathbf{/}\mathbf{3}}{\mathbf{-}}{{\mathbit{c}}}_{{\mathbf{3}}}\frac{\mathbf{Z}\mathbf{\left(}\mathbf{Z}\mathbf{-}\mathbf{1}\mathbf{\right)}}{{\mathbf{A}}^{\mathbf{1}\mathbf{/}\mathbf{3}}}{\mathbf{-}}{{\mathbit{c}}}_{{\mathbf{4}}}\frac{{\mathbf{\left(}\mathbf{N}\mathbf{-}\mathbf{Z}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{A}}$ .

Where, ${c}_{1}=15.8\text{MeV,}{c}_{2}=17.8\text{MeV,}{c}_{3}=0.71\text{MeV,}{c}_{4}=23.7\text{MeV}$.

$A\to$Mass Number, $Z\to$ Atomic Number and $Z\to$ Number of neutrons.

## Step 3: Calculation for the factor that differs in nuclear binding of oxygen and nitrogen

(a)

Volume term, ${c}_{1}A$ and surface term $\left[-{c}_{2}{A}^{2/3}\mid \right$ is depend upon mass number A.

Since mass number is same in both oxygen-15 and nitrogen-15 . So volume term and surface term is same in both.

The coulomb term and the last term, $\left[-{c}_{3}\frac{Z\left(Z-1\right)}{{A}^{1/3}}\right]$ and asymmetry term $\left[-{c}_{4}\frac{{\left(N-Z\right)}^{2}}{A}\right]$ may be different because in both oxygen- 15 and nitrogen-15 , atomic number Z and number of neutrons N is different.

## Step 4: Calculation to find which isobar is more tightly bound

(b)

Volume term, ${c}_{1}A$ and surface term $\left[-{c}_{2}{A}^{2/3}\right]$ is depend upon mass number .

Since mass number is same in both oxygen-15 and nitrogen- 15. So volume term and surface term is same in both.

The coulomb term for oxygen-15 : $\left[-{c}_{3}\frac{Z\left(Z-1\right)}{{A}^{1/3}}\right]=\left[-\left(0.71\text{MeV}\right)\frac{\left(8\right)\left(7\right)}{{\left(15\right)}^{1/3}}\right]=-16.12\text{MeV}$

Asymmetry term for oxygen-15 : $\left[-c4\frac{{\left(N-Z\right)}^{2}}{A}\right]=\left[-\left(23.7\text{MeV}\right)\frac{{\left(7-8\right)}^{2}}{\left(15\right)}\right]=-1.58\text{MeV}$

The coulomb term for nitrogen-15 : $\left[-{c}_{3}\frac{Z\left(Z-1\right)}{{A}^{1/3}}\right]=\left[-\left(0.71\text{MeV}\right)\frac{\left(7\right)\left(6\right)}{{\left(15\right)}^{1/3}}\right]=-12.09\text{MeV}$

Asymmetry term for nitrogen-15 : $\left[-{c}_{3}\frac{Z\left(Z-1\right)}{{A}^{1/3}}\right]=\left[-\left(0.71\text{MeV}\right)\frac{\left(7\right)\left(6\right)}{{\left(15\right)}^{1/3}}\right]=-12.09\text{MeV}$

All term are same except coulomb term of semi empirical binding energy of both isobars.

Since The coulomb term for oxygen-15 more negative than coulomb term for nitrogen- 15, so the binding energy of nitrogen-15 is more and it is more tightly bound.

## Step 5: Calculation to show that nitrogen is more tightly bound

(c)

Half-life decay mode of oxygen-15 is 122.2 second.

Half – life decay mode of nitrogen- 15 is more than 122.2 second.

Half-life decay mode of nitrogen-15 is more than oxygen-15, so the half-life decay mode support the nitrogen's nucleon's tightly bound statement that nitrogen-15 is more tightly bound.

Yes, half-life decay mode support the nitrogen's nucleons tightly bound statement.

## Step 6: Calculation for the binding energy of oxygen and nitrogen

(d)

The binding energy for oxygen- 15 is, $\text{BE}=\left(8{M}_{\text{H}}+7m-{M}_{\text{s}s}N\right){c}^{2}$ .

Substitute $1.007825a$ for ${M}_{\mathbf{H}},1.008665$ u for ${m}_{\text{a}}$ and $15.003065$ a for ${M}_{\text{k}}$ o in above equation.

$\text{BE}=\left[8\left(1007825\text{u}\right)+7\left(1008665\text{u}\right)-\left(15.003065\text{u}\right)\right]\left(931.5\text{MeV}/\text{u}\right)\phantom{\rule{0ex}{0ex}}\text{BE}=111.96\text{MeV}$

The binding energy for nitrogen-15 : $\text{BE}=\left(7{M}_{\text{H}}+8{m}_{\text{a}}-{M}_{\text{g}5}\text{N}\right){c}^{2}$

Substitute $1.007825\text{n}$ for ${M}_{\mathbf{H}},1.008665\text{u}$ for ${m}_{\text{n}}$ and $15.000108\text{u}$ for ${M}_{\text{k}}\text{n}$ in above equation.

$\text{BE}=\left[7\left(1.007825\text{u}\right)+8\left(1.008665\text{u}\right)-\left(15.000108\text{u}\right)\right]\left(9315\text{MeV}/\text{u}\right)\phantom{\rule{0ex}{0ex}}\text{BE}=115.49\text{MeV}$

The difference is given as:

of nitrogen- $15-\text{BE}$ of oxygen $-15=115.49\text{MeV}-111.96\text{MeV}$ .

(e)

The semi empirical binding energy formula is $BE={c}_{1}A-{c}_{2}{A}^{2/3}-{c}_{3}\frac{Z\left(Z-1\right)}{{A}^{2/3}}-{c}_{4}\frac{{\left(N-Z\right)}^{2}}{A}$.

The binding energy for oxygen- 15 is, $\text{BE}=\left(8{M}_{\text{H}}+7m-{M}_{\text{s}s}N\right){c}^{2}$.

Substitute $1.007825a$ for ${M}_{\mathbf{H}},1.008665$ u for ${m}_{\text{a}}$ and $15.003065$ a for ${M}_{\text{k}}$ o in above equation.

$\text{BE}=\left[8\left(1007825\text{u}\right)+7\left(1008665\text{u}\right)-\left(15.003065\text{u}\right)\right]\left(931.5\text{MeV}/\text{u}\right)\phantom{\rule{0ex}{0ex}}\text{BE}=111.96\text{MeV}$

The binding energy for nitrogen-15 : $\text{BE}=\left(7{M}_{\text{H}}+8{m}_{\text{a}}-{M}_{\text{g}5}\text{N}\right){c}^{2}$

Substitute $1.007825\text{n}$ for ${M}_{\mathbf{H}},1.008665\text{u}$ for ${m}_{\text{n}}$ and $15.000108\text{u}$ for ${M}_{\text{k}}\text{n}$ in above equation.

$\text{BE}=\left[7\left(1.007825\text{u}\right)+8\left(1.008665\text{u}\right)-\left(15.000108\text{u}\right)\right]\left(9315\text{MeV}/\text{u}\right)\phantom{\rule{0ex}{0ex}}\text{BE}=115.49\text{MeV}$

The difference is given as:

BE of nitrogen- 15-BE of oxygen $-15=115.49\text{MeV}-111.96\text{MeV}$ $=3.53\text{MeV}$.