Answer
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Hint: To solve this question, we have to remember some basic concepts of nuclear reactions. Mass defect of a nucleus is always less than the total mass of the constituent nucleons. If a certain number of neutrons and protons are brought together to form nucleus, the mass of the nucleons forming the nucleus decreases or defects.
Complete answer:
The difference between the sum of the masses of the nucleons constituting a nucleus and the rest mass of the nucleus is known as mass defect. It is denoted by $\Delta m$.
Let us calculate the mass defect in case of the nucleus of an atom ${}_Z{X^A}$. The nucleus of such an atom contains Z protons and ( A – Z ) neutrons.
Mass of protons = ${m_n}$
Mass of proton = ${m_p}$
Mass of nucleus = ${m_N}$
Therefore,
Mass of the constituting nucleons = $Z{m_p} + \left( {A - Z} \right){m_n}$
If ${m_N}\left( {{}_Z{X^A}} \right)$ is mass of the nucleus of the atom ${}_Z{X^A}$, then the mass defect is given by,
$ \Rightarrow \Delta m = \left[ {Z{m_p} + \left( {A - Z} \right){m_n}} \right] - {m_N}\left( {{}_Z{X^A}} \right)$
Binding energy: when a nucleus is formed from the free nucleons, the decrease in mass of the nucleons is released as equivalent energy. The energy equivalent to mass defect is used in binding the nucleons and is called the binding energy.
If $\Delta m$ is the mass defect of a nucleus, then according to Einstein’s mass energy relation,
Binding energy = $\Delta m{c^2}$( in joule ).
Binding energy = $\left( {\left[ {Z{m_p} + \left( {A - Z} \right){m_n}} \right] - {m_N}\left( {{}_Z{X^A}} \right)} \right){c^2}$( in joule ).
Binding energy per nucleon is the average energy required to extract one nucleon from the nucleus. It is obtained by dividing the binding energy of the nucleus by the number of nucleons it contains.
Thus,
Binding energy per nucleon = $\dfrac{{{\text{Binding energy}}}}{{\text{A}}} = \dfrac{{\Delta m{c^2}}}{A}$
Note:
Binding energy is sometimes considered as a negative quantity, since it is an energy deficit. Binding energy of the nucleus may be termed as the work done against the binding force to pull the nucleons apart. Mass defect and binding energy of the hydrogen nucleus is zero. It is because there is only one nucleon.
Complete answer:
The difference between the sum of the masses of the nucleons constituting a nucleus and the rest mass of the nucleus is known as mass defect. It is denoted by $\Delta m$.
Let us calculate the mass defect in case of the nucleus of an atom ${}_Z{X^A}$. The nucleus of such an atom contains Z protons and ( A – Z ) neutrons.
Mass of protons = ${m_n}$
Mass of proton = ${m_p}$
Mass of nucleus = ${m_N}$
Therefore,
Mass of the constituting nucleons = $Z{m_p} + \left( {A - Z} \right){m_n}$
If ${m_N}\left( {{}_Z{X^A}} \right)$ is mass of the nucleus of the atom ${}_Z{X^A}$, then the mass defect is given by,
$ \Rightarrow \Delta m = \left[ {Z{m_p} + \left( {A - Z} \right){m_n}} \right] - {m_N}\left( {{}_Z{X^A}} \right)$
Binding energy: when a nucleus is formed from the free nucleons, the decrease in mass of the nucleons is released as equivalent energy. The energy equivalent to mass defect is used in binding the nucleons and is called the binding energy.
If $\Delta m$ is the mass defect of a nucleus, then according to Einstein’s mass energy relation,
Binding energy = $\Delta m{c^2}$( in joule ).
Binding energy = $\left( {\left[ {Z{m_p} + \left( {A - Z} \right){m_n}} \right] - {m_N}\left( {{}_Z{X^A}} \right)} \right){c^2}$( in joule ).
Binding energy per nucleon is the average energy required to extract one nucleon from the nucleus. It is obtained by dividing the binding energy of the nucleus by the number of nucleons it contains.
Thus,
Binding energy per nucleon = $\dfrac{{{\text{Binding energy}}}}{{\text{A}}} = \dfrac{{\Delta m{c^2}}}{A}$
Note:
Binding energy is sometimes considered as a negative quantity, since it is an energy deficit. Binding energy of the nucleus may be termed as the work done against the binding force to pull the nucleons apart. Mass defect and binding energy of the hydrogen nucleus is zero. It is because there is only one nucleon.
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