The binding energy of nucleons in a nucleus can be affected by the pairwise Coulomb repulsion. Assume that all nucleons are uniformly distributed inside the nucleus. Let the binding energy of a proton be E_b^p and the binding energy of a neutron be E_b^n in the nucleus. Which of the following statement(s) is(are) correct?

E_b^p-E_b^n is proportional to A^-\frac13 where A is the mass number of the nucleus., E_b^p increases if the nucleus undergoes a beta decay emitting a positron.

E_b^p-E_b^n is proportional to Z(Z-1) where Z is the atomic number of the nucleus.

E_b^p-E_b^n is proportional to A^-\frac13 where A is the mass number of the nucleus.

E_b^p increases if the nucleus undergoes a beta decay emitting a positron.

Atoms and Nuclei PYQ — JEE Advanced 2022 · Shift 1 · Q50 with Solution

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Correct answer: B, D

Idea: difference between proton and neutron binding energies

A neutron does not experience Coulomb repulsion, but a proton does. So the difference $E_b^p-E_b^n$ comes from the Coulomb energy contribution.

If the nucleus has $Z$ protons and all nucleons are uniformly distributed, then the total electrostatic potential energy of the protons is

U_C = \frac{3}{5}\frac{1}{4\pi\varepsilon_0}\frac{Z(Z-1)e^2}{R}

where nuclear radius

R=r_0 A^{1/3}.

Hence

U_C \propto \frac{Z(Z-1)}{A^{1/3}}.

Coulomb contribution per proton

The average Coulomb energy associated with one proton is of the order

\frac{U_C}{Z} \propto \frac{Z-1}{A^{1/3}}.

Equivalently, the potential inside a uniformly charged sphere is proportional to total charge divided by radius, so for a proton inside the nucleus,

\Delta E \sim \frac{Ze^2}{R} \propto \frac{Z}{A^{1/3}}.

Thus the proton is less bound than the neutron due to Coulomb repulsion. Therefore,

E_b^p - E_b^n < 0.

So option C is false.

Check option A

Option A says:

E_b^p-E_b^n \propto Z(Z-1)

But the difference for one proton relative to one neutron should scale like Coulomb potential per proton, i.e. roughly

E_b^p-E_b^n \propto -\frac{Z-1}{R} \propto -(Z-1)A^{-1/3}.

It is not proportional to the total pair count $Z(Z-1)$ .

So A is false.

Check option B

Since

R\propto A^{1/3},

the Coulomb effect per proton varies as

E_b^p-E_b^n \propto -\frac{1}{R} \propto -A^{-1/3}

(for fixed $Z$ -dependence separately). Thus the dependence on $A$ is indeed proportional to $A^{-1/3}$ .

So B is true.

Check option C

Because proton suffers Coulomb repulsion, it is less tightly bound than neutron:

E_b^p < E_b^n

Hence

E_b^p-E_b^n < 0.

So C is false.

Check option D

In $\beta^+$ decay,

p \to n + e^+ + \nu_e

inside the nucleus, so the daughter nucleus has

Z \to Z-1, \qquad A = \text{constant}.

Since Coulomb repulsion decreases, the remaining protons are more tightly bound. Therefore $E_b^p$ increases.

So D is true.

Final derived answer

Correct options are:

\boxed{\text{B, D}}

Comparison with stored answer

Stored answer: A, B, D

My derivation shows that A is not correct because $E_b^p-E_b^n$ for an individual nucleon should scale with Coulomb potential per proton, not total Coulomb energy of the nucleus. Therefore I disagree with the stored answer.

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View written solutionFree

Correct answer: B, D

Idea: difference between proton and neutron binding energies

A neutron does not experience Coulomb repulsion, but a proton does. So the difference $E_b^p-E_b^n$ comes from the Coulomb energy contribution.

If the nucleus has $Z$ protons and all nucleons are uniformly distributed, then the total electrostatic potential energy of the protons is

U_C = \frac{3}{5}\frac{1}{4\pi\varepsilon_0}\frac{Z(Z-1)e^2}{R}

where nuclear radius

R=r_0 A^{1/3}.

Hence

U_C \propto \frac{Z(Z-1)}{A^{1/3}}.

Coulomb contribution per proton

The average Coulomb energy associated with one proton is of the order

\frac{U_C}{Z} \propto \frac{Z-1}{A^{1/3}}.

Equivalently, the potential inside a uniformly charged sphere is proportional to total charge divided by radius, so for a proton inside the nucleus,

\Delta E \sim \frac{Ze^2}{R} \propto \frac{Z}{A^{1/3}}.

Thus the proton is less bound than the neutron due to Coulomb repulsion. Therefore,

E_b^p - E_b^n < 0.

So option C is false.

Check option A

Option A says:

E_b^p-E_b^n \propto Z(Z-1)

But the difference for one proton relative to one neutron should scale like Coulomb potential per proton, i.e. roughly

E_b^p-E_b^n \propto -\frac{Z-1}{R} \propto -(Z-1)A^{-1/3}.

It is not proportional to the total pair count $Z(Z-1)$ .

So A is false.

Check option B

Since

R\propto A^{1/3},

the Coulomb effect per proton varies as

E_b^p-E_b^n \propto -\frac{1}{R} \propto -A^{-1/3}

(for fixed $Z$ -dependence separately). Thus the dependence on $A$ is indeed proportional to $A^{-1/3}$ .

So B is true.

Check option C

Because proton suffers Coulomb repulsion, it is less tightly bound than neutron:

E_b^p < E_b^n

Hence

E_b^p-E_b^n < 0.

So C is false.

Check option D

In $\beta^+$ decay,

p \to n + e^+ + \nu_e

inside the nucleus, so the daughter nucleus has

Z \to Z-1, \qquad A = \text{constant}.

Since Coulomb repulsion decreases, the remaining protons are more tightly bound. Therefore $E_b^p$ increases.

So D is true.

Final derived answer

Correct options are:

\boxed{\text{B, D}}

Comparison with stored answer

Stored answer: A, B, D

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