Question

The mass of ${}_{\text{3}}{\text{L}}{{\text{i}}^{\text{7}}}$ nucleus is $0.042\,\,a.m.u$ less than the sum of masses of its nucleons. Find the binding energy per nucleon.

Answer 1

Hint: Here in this question we will first find the binding energy by using the formula \[{\text{B}}{\text{.E}}{\text{.}} = \left[ {Zmp + \left( {A - Z} \right){m_n} - {M_N}} \right]{C^2}\] and after finding it, then we have to find the binding energy per nucleon by using the formula, ${\text{Binding energy per nucleon = }}\dfrac{{{\text{Total binding energy of nucleus}}}}{{{\text{Number of nucleons in the nucleus }}}}$ and on substituting the values and solving it we will get the answer.
Formula used:
Binding energy,
\[{\text{B}}{\text{.E}}{\text{.}} = \left[ {Zmp + \left( {A - Z} \right){m_n} - {M_N}} \right]{C^2}\]
Where,
\[\Delta m = \left[ {Zmp + \left( {A - Z} \right){m_n} - {M_N}} \right]\]
Here,
$\Delta m$ , is mass defect
So, Binding energy, ${\text{B}}{\text{.E}}{\text{.}} = \left( {\Delta m} \right){c^2}$
In this formula,
$Z$ , is the atomic number
$A$ , is the mass number
${m_p}$ , is the mass of proton
${m_n}$ , is the mass of neutron,
${m_N}$ , is the mass of nucleus \[\left( {_Z{X^A}} \right)\]
$C$ , is the velocity of light

Complete Step by Step Answer:
According to question we know that the mass of ${}_{\text{3}}{\text{L}}{{\text{i}}^{\text{7}}}$ nucleus is \[0.042a.m.u.\] less than sum of masses of its nucleons. Which means that the mass defect \[\left( {\Delta m} \right)\] is in \[a.m.u.\] revised in
${\text{MeV}}$ can be calculated directly by multiplying mass defect with \[931\left( {MeV} \right)\]
So, the binding energy
$ \Rightarrow \left( {B.E} \right) = \left( {\Delta m} \right) \times 931MeV$
Now on substituting the values, we get
$ \Rightarrow 0.042 \times 931$
And on solving it, we get
$ \Rightarrow 39.102MeV$
To calculate the binding energy per nucleon, we will use the formula
$
  {\text{Binding energy per nucleon = }}\dfrac{{{\text{Total binding energy of nucleus}}}}{{{\text{Number of nucleons in the nucleus }}}} \\
    \\
$
In nucleus${{\text{ }}_{\text{3}}}{\text{L}}{{\text{i}}^{\text{7}}}$ , Number of nucleons will be equal to $7$
As in a nucleus, Number of nucleons will be equal to the number of protons and the addition of the number of neutrons in it.
As in $_{\text{3}}{\text{L}}{{\text{i}}^{\text{7}}}$ , Number of protons is $3$ , Atomic Number is also $3$ and Mass number will be equal to the number of neutron
$ \Rightarrow 3 + {\text{Number}}\,\,{\text{of}}\,\,{\text{neutrons}} = 7$ and the number of neutrons will be equal to $7$
So, Binding energy per nucleon
\[ \Rightarrow \dfrac{{39.102}}{7}\]
And on solving it we get
\[ \Rightarrow 5.58MeV \approx 5.6MeV\]

Therefore, the binding energy per nucleon is equal to \[5.6MeV\] .

Note: Binding energy of a nucleus is the energy with which nucleons are bound in the nucleus. It is measured by the work required to be done to separate the nucleons an infinite distance apart from the nucleons, so that they may not interact with each other.