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Q26E

Expert-verifiedFound in: Page 1

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 $3.53\text{MeV}$.

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

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, $N=A-Z=15-7=8$.

**Semi empirical binding Energy is given by the expression, **

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

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.

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

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

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

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

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

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.

(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.

(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.

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

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.

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

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 $BE={c}_{1}A-{c}_{2}{A}^{2/3}-{c}_{3}\frac{Z(Z-1)}{{A}^{2/3}}-{c}_{4}\frac{{(N-Z)}^{2}}{A}$.

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.

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

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.

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

The difference is given as:

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

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