Question

What is the binding energy per nucleon of the Rutherfordium isotope \[_{104}^{259}Rf\] ? Here are some atomic masses and the neutron mass.
\[_{104}^{259}Rf\]\[259.10563u\] \[^1H\] \[1.007825u\] \[n\]\[1.008655u\]

Answer 1

Hint: As we know that nucleus contain protons and electrons. Binding energy is the amount of energy required to separate a particle from a system of particles.
Binding energy =change in mass\[ \times \]\[{c^2}\]
Here c \[ \to \] Speed of light in vacuum
C = \[2.9979 \times {10^8} \approx 3 \times {10^8}\]\[m\]⁄\[s\]

Formula used:
The binding energy of nucleus is given by
$\Delta$ \[{E_{be}}\] = \[\sum {\left( {m{c^2}} \right)} \]- \[M{c^2}\] Where \[\sum {\left( {m{c^2}} \right)} \]represents total mass energy of protons and neutrons.

Complete step by step solution:
The binding energy of nucleus is given by
$\Delta$\[{E_{be}}\] = \[\sum {\left( {m{c^2}} \right)} \]- \[M{c^2}\]
Where \[\sum {\left( {m{c^2}} \right)} \]represents total mass energy of protons and neutrons, the number of protons is represented by\[z\]and the number of neutrons is represented by\[A - Z\], \[{m_p}\]represents mass of proton and \[{m_n}\] represents mass of neutrons.
\[\sum {\left( {m{c^2}} \right)} \]=\[\left( {Z{m_p} + \left( {A - Z} \right){m_n}} \right){c^2}\]
Substitute into above equation
$\Delta$\[{E_{be}}\]= \[\left( {Z{m_p} + \left( {A - Z} \right){m_n} - M} \right){c^2}\]
Given that
For \[_{104}^{259}Rf\], Mass = \[259.10563\]\[u\], \[{m_p}\]=\[1.007825u\], \[{m_n}\]=\[1.008655u\]
Substitute these values, \[\dfrac{{{E_{be}}}}{{number of nucleons}}\]
$\Delta$\[{E_{be}}\]=\[\left( {\left( {104} \right)\left( {1.007825u} \right) + \left( {259 - 104} \right)\left( {1.008665u} \right) - 259.1053u} \right)\]\[{c^2}\]= \[\left( {2.05124u} \right)\]\[{c^2}\]
=\[\dfrac{{1910.722437MeV}}{{259}}\]
\[1u\]= \[931.49401MeV\]⁄\[{c^2}\]
$\Delta$\[{E_{be}}\]=(\[2.051245 \times 931.494013MeV\] ⁄\[{c^2}\])\[{c^2}\]=\[1910.722437MeV\]
The binding energy per nucleon \[ = \dfrac{{energy}}{{number\,of\,nucleons}}\]
$\Delta$\[{E_{be,n}}\] = \[\dfrac{{1910.722437MeV}}{{259}}\]=\[7.3773MeV\]

Note:
The binding energy of nuclei is always positive. Binding energy plays an important role in nuclear fusion. The nuclear binding energy depends upon the asymmetry between the number of protons and neutrons and also depends upon the coulomb repulsion force. Nuclear fission and nuclear fusion are the phenomena that depend upon the binding energy. In fact, without binding energy nuclei study is impossible.