91Ó°ÊÓ

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, $c_1=15.8\text{MeV,}{c}_2=17.8\text{MeV,}{c}_3=0.71\text{MeV,}{c}_4=23.7\text{MeV}$.

$A→$Mass Number, $Z→$Atomic Number and $Z→$Number of neutrons.

(a)

Volume term, $c_{1}A$and surface term $\left[-c_2{A}^{2/3}âˆ\pounds \right.$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, $\text{BE}=\left(8M_{\text{H}}+7m-M_{\text{s}s}N\right)c^2$.

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.

$$\begin{gathered} \text{BE}=\left[8\right(1007825\text{u})+7(1008665\text{u})-(15.003065\text{u}\left)\right](931.5\text{MeV}/\text{u}) \\ \text{BE}=111.96\text{MeV} \end{gathered}$$

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

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.

$$\begin{gathered} \text{BE}=\left[7\right(1.007825\text{u})+8(1.008665\text{u})-(15.000108\text{u}\left)\right](9315\text{MeV}/\text{u}) \\ \text{BE}=115.49\text{MeV} \end{gathered}$$

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, $\text{BE}=\left(8M_{\text{H}}+7m-M_{\text{s}s}N\right)c^2$.

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.

$$\begin{gathered} \text{BE}=\left[8\right(1007825\text{u})+7(1008665\text{u})-(15.003065\text{u}\left)\right](931.5\text{MeV}/\text{u}) \\ \text{BE}=111.96\text{MeV} \end{gathered}$$

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

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.

$$\begin{gathered} \text{BE}=\left[7\right(1.007825\text{u})+8(1.008665\text{u})-(15.000108\text{u}\left)\right](9315\text{MeV}/\text{u}) \\ \text{BE}=115.49\text{MeV} \end{gathered}$$

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}$.