Question

The mass of a $ {}_3^7Li $ nucleus is $ 0.042u $ less than the sum of the masses of all its nucleons. The binding energy per nucleon of $ {}_3^7Li $ nucleus is nearly:
A. $ 23MeV $
B. $ 46MeV $
C. $ 5.6MeV $
D. $ 3.9MeV $

Answer 1

Hint : Using the information given in this question, we have to substitute suitable values in the formula of binding energy per nucleon. After that we have to convert the energy value into $ MeV $ by multiplying the value with a factor $ 931.5 $ .

Formulas used:
 $ {E_B} = \left[ {Z{m_p} + \left( {A - Z} \right){m_n} - {M_n}} \right]{c^2} $
 $ {E_B} = \left[ {Z{m_p} + \left( {A - Z} \right){m_n} - {M_n}} \right] \times 931.5MeV $
Where $ {m_p} $ is the mass of a proton and $ {m_n} $ is the mass of a neutron.
 $ {M_n} $ is the mass of a nucleus, $ c $ is the speed of light in vacuum, $ A $ is the mass number and $ Z $ is the atomic number.
 $ {f_B} = \dfrac{{{E_B}}}{A} = \left[ {\dfrac{{Z{m_p} + \left( {A - Z} \right){m_n} - {M_n}}}{A}} \right]{c^2} $
 $ {f_B} = $ $ \dfrac{{{E_B}}}{A} = \dfrac{{Z{m_p} + \left( {A - Z} \right){m_n} - {M_n}}}{A} \times 931.5MeV $ per nucleon
Where $ {f_B} $ is the binding energy per nucleon.

Complete step by step answer
Binding energy of a nucleus is defined as the minimum amount of energy which must be supplied to the nucleus to break it into its constituent nucleons. Now, the specific binding energy is defined as the average binding energy per nucleon.
The mass of any stable nucleus is found to be less than the sum of the mass of all the protons and the neutrons inside the nucleus. This lost mass in the form of energy keeps the nucleus together and is termed as binding energy.
If $ Z $ is the atomic number and $ A $ is the mass number then the binding energy is given as,
 $ {E_B} = \left[ {Z{m_p} + \left( {A - Z} \right){m_n} - {M_n}} \right]{c^2} $
Where $ {m_p} $ is the mass of a proton and $ {m_n} $ is the mass of a neutron.
 $ {M_n} $ is the mass of a nucleus and $ c $ is the speed of light in vacuum.
The expression for the binding energy per nucleon or the specific binding energy is given as
 $ {f_B} = \dfrac{{{E_B}}}{A} = \left[ {\dfrac{{Z{m_p} + \left( {A - Z} \right){m_n} - {M_n}}}{A}} \right]{c^2} $
First, let us calculate the sum of the masses of all its nucleons which is
 $ \begin{gathered}
  Z{m_p} + \left( {A - Z} \right){m_n} = \left( {3 \times 1.007276} \right) + \left( {4 \times 1.008665} \right) \\
   = 7.056488u \\
\end{gathered} $
Where $ {m_p} = 1.007276 $ and $ {m_n} = 1.008665u $
Now, the mass of the nucleus is
 $ {M_n} = 7.056488 - 0.042 = 7.014488u $
So, the Binding energy per nucleon is given as,
 $ \begin{gathered}
  {f_B} = \dfrac{{7.056488 - 7.01448}}{7} = \dfrac{{0.042}}{7} = 0.006u \\
   \Rightarrow {f_B} = 0.006 \times 931.5MeV = 5.589MeV \\
\end{gathered} $
As mass is expressed in $ a.m.u $ (atomic mass unit), in the unit of energy ( $ 1a.m.u = 931.5MeV $ )
1. Binding energy can be expressed as, $ {E_B} = \left[ {Z{m_p} + \left( {A - Z} \right){m_n} - {M_n}} \right] \times 931.5MeV $
2. Binding energy per nucleon is given as, $ \dfrac{{{E_B}}}{A} = \dfrac{{Z{m_p} + \left( {A - Z} \right){m_n} - {M_n}}}{A} \times 931.5MeV $ per nucleon.
Therefore, the correct answer is option C.

Note
When a nucleus is stable, energy is required from outside to disrupt it into its constituents separately, whereas when a nucleus is unstable, it will disintegrate by itself. So binding energy of a nucleus gives us information about its stability.