1. Use the semi-empirical mass formula ideas

For a nucleus, the binding energy contains terms like:

$B = a_v A - a_s A^{2/3} - a_c \frac{Z(Z-1)}{A^{1/3}} + \cdots$

If we talk about binding energy per nucleon, divide by $A$:

$\frac{B}{A} = a_v - a_s A^{-1/3} - a_c \frac{Z(Z-1)}{A^{4/3}} + \cdots$

Now check each statement.

2. Statement A

It says the surface energy per nucleon is

$b_s = -a_1 A^{2/3}$

But surface energy itself is proportional to surface area:

$E_s \propto -A^{2/3}$

So surface energy per nucleon is

$b_s = \frac{E_s}{A} \propto -A^{-1/3}$

Hence statement A is false.

3. Statement B

It says Coulomb contribution to the binding energy is

$b_c = -a_2 \frac{Z(Z-1)}{A^{4/3}}$

But the Coulomb term in the total binding energy is

$E_c = -a_c \frac{Z(Z-1)}{A^{1/3}}$

The given power $A^{-4/3}$ corresponds to binding energy per nucleon, not total binding energy.

Therefore B is false as stated.

4. Statement C

It says volume energy is

$b_v = a_3 A$

The volume term in total binding energy is indeed proportional to $A$:

$E_v = a_v A$

So C is true.

5. Statement D

Surface effect reduces binding because nucleons at the surface have fewer neighbors. Since surface area of nucleus is proportional to

$R^2 \propto (A^{1/3})^2 = A^{2/3}$

Decrease in binding energy due to surface is proportional to surface area.

So D is true.

6. Statement E

In estimating surface energy, the idea used is that each nucleon interacts only with near neighbors due to saturation of nuclear force. One often says each nucleon interacts with a limited number of neighboring nucleons, but the fixed number 12 is not a standard assumption of the liquid drop derivation here.

Hence E is not taken as correct in this context.

7. Final evaluation

• A: False
• B: False
• C: True
• D: True
• E: False

Therefore the correct choice is:

$\boxed{\text{C, D only}}$

So the correct option is A.