Question

Find the binding energy per nucleon of a rutherfordium isotope ${}_{104}^{259}Rf$ . Here are some atomic masses and neutron masses.
${}_{104}^{259}Rf \to 259 \cdot 10563{\text{u}}$
${}^1H \to 1 \cdot 007825{\text{u}}$
$n \to 1 \cdot 008665{\text{u}}$

Answer 1

A nucleus contains protons and neutrons. The binding energy of a nucleus is defined as the difference between the mass-energy of the nucleons present in the given nucleus and the mass-energy of the given nucleus The binding energy per nucleon of the given nucleus will be the binding energy divided by the number of protons and neutrons present in it.

Formula used:
-The binding energy of a nucleus is given by, $\Delta {E_{be}} = \sum {\left( {m{c^2}} \right)} - M{c^2}$ where $\sum {\left( {m{c^2}} \right)}$ is the total mass-energy of all the nucleons in the nucleus, $M$ is the atomic mass of the nucleus and $c$ is the speed of light.

Complete step by step answer.
Step 1: List the parameters given in the question.
We have to obtain the binding energy per nucleon of the given rutherfordium isotope ${}_{104}^{259}Rf$ .
The mass number of the isotope is $A = 259$ and the atomic number of the isotope is $Z = 104$ .
The mass number is the sum of the number of protons $Z$ and the number of neutrons $N$ in the nucleus i.e., $A = Z + N$ .
So the number of neutrons in the rutherfordium nucleus will be $N = A - Z = 259 - 104 = 155$ .
The atomic mass of the given rutherfordium isotope is given to be $M = 259 \cdot 10563{\text{u}}$ .
The atomic mass of a hydrogen nucleus is given to be ${M_H} = 1 \cdot 007825{\text{u}}$ .
The mass of a neutron is given to be ${m_n} = 1 \cdot 008665{\text{u}}$ .
Step 2: Express the relation for the binding energy of the given rutherfordium isotope.
The binding energy of the given rutherfordium isotope can be expressed as
$\Delta {E_{be}} = \sum {\left( {m{c^2}} \right)} - M{c^2}$ ----------- (1)
where $\sum {\left( {m{c^2}} \right)}$ is the total mass-energy of all the nucleons in the nucleus, $M$ is the atomic mass of the given nucleus and $c$ is the speed of light.
Now the total mass-energy will be $\sum {\left( {m{c^2}} \right)} = \left( {Z{m_p} + N{m_n}} \right){c^2}$ --------- (2)
Substituting for $Z = 104$ , $N = 155$ , ${m_p} = 1 \cdot 007825{\text{u}}$ and ${m_n} = 1 \cdot 008665{\text{u}}$ in equation (2) we get, $\sum {\left( {m{c^2}} \right)} = \left[ {\left( {104 \times 1 \cdot 007825} \right) + \left( {155 \times 1 \cdot 008665} \right)} \right]{c^2} = 261 \cdot 156875{c^2}$
Thus the total mass-energy of the nucleons in the given isotope is $\sum {\left( {m{c^2}} \right)} = 261 \cdot 156875{c^2}$ .
Now substituting for $\sum {\left( {m{c^2}} \right)} = 261 \cdot 156875{c^2}$ and $M = 259 \cdot 10563{\text{u}}$ in equation (1) we get, $\Delta {E_{be}} = 261 \cdot 156875{c^2} - 259 \cdot 10563{c^2} = \left( {2 \cdot 051245{\text{u}}} \right){c^2}$
Since $1{\text{u}} = \dfrac{{931 \cdot 494013{\text{MeV}}}}{{{c^2}}}$ , the binding energy of the given nucleus will be $\Delta {E_{be}} = \dfrac{{\left( {2 \cdot 051245 \times 931 \cdot 494013{\text{MeV}}} \right){c^2}}}{{{c^2}}} = 1910 \cdot 722437{\text{MeV}}$
Thus the binding energy of the rutherfordium nucleus is $\Delta {E_{be}} = 1910 \cdot 722437{\text{MeV}}$ .
Step 3: Express the binding energy per nucleon of the given nucleus.
The total number of nucleons in the given rutherfordium nucleus is given by the mass number of the nucleus $A = 259$ .
So the binding energy per nucleon of the rutherfordium isotope will be
$\Delta {E_{be/nucleons}} = \dfrac{{\Delta {E_{be}}}}{A}$ --------- (3)
Substituting for $\Delta {E_{be}} = 1910 \cdot 722437{\text{MeV}}$ and $A = 259$ in equation (3) we get, $\Delta {E_{be/nucleons}} = \dfrac{{1910 \cdot 722437{\text{MeV}}}}{{259}} = 7 \cdot 3773{\text{MeV}}$
Thus we obtain the binding energy per nucleon of the given rutherfordium nucleus as $\Delta {E_{be/nucleons}} = 7 \cdot 3773{\text{MeV}}$ .

Note: A hydrogen nucleus consists only of a proton. So the atomic mass of the hydrogen nucleus will be the mass of the proton present in it i.e., ${M_H} = {m_p}$ . Thus we substituted ${m_p} = 1 \cdot 007825{\text{u}}$ in equation (2). The binding energy of a nucleus is usually expressed in mega electron-volts so we apply the unit conversion from ${\text{u}}$ to ${\text{MeV}}$ .