1. Set up the 1D elastic collision

Let the alpha-particle have mass $m$ and initial velocity $u$. The nucleus is initially at rest and has mass $M$.

After collision, the alpha-particle is scattered directly backwards, so its final velocity is negative.

For a 1-dimensional elastic collision with target initially at rest, the final velocity of the projectile is

$v_1 = \frac{m-M}{m+M}u$

Since it goes backward, we must have $M>m$ so that $v_1<0$.

2. Use the kinetic energy loss condition

Initial kinetic energy of alpha-particle:

$K_i = \frac12 mu^2$

It loses $64\%$ of its initial kinetic energy, so it retains $36\%$:

$K_f = 0.36 K_i$

Thus,

$\frac{\frac12 m v_1^2}{\frac12 m u^2} = 0.36$

$\Rightarrow \frac{v_1^2}{u^2} = 0.36$

$\Rightarrow \left(\frac{m-M}{m+M}\right)^2 = 0.36 = \left(\frac35\right)^2$

So,

$\left|\frac{m-M}{m+M}\right| = \frac35$

Because the particle goes backward, $v_1<0$, hence

$\frac{m-M}{m+M} = -\frac35$

3. Solve for $M$

$\frac{m-M}{m+M} = -\frac35$

Cross-multiplying:

$5(m-M) = -3(m+M)$

$5m - 5M = -3m - 3M$

$8m = 2M$

$M = 4m$

4. Check with options

The mass of the nucleus is

$\boxed{4m}$

So the correct option is B.

5. Comparison with stored answer

Stored correct answer: B

Derived answer: B

They match.