The De Broglie wavelength of a particle of mass $m$ and kinetic energy $K$ is:

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m K}}$$

For particles having the identical kinetic energy $K$:

$$\lambda \propto \frac{1}{\sqrt{m}}$$

Comparing the masses of the given particles:

• Electron: $m_e \approx 9.1 \times 10^{-31}\,\text{kg}$
• Proton: $m_p \approx 1\,\text{u}$
• Deuteron: $m_d \approx 2\,\text{u}$
• $\alpha$-particle: $m_\alpha \approx 4\,\text{u}$

Since the $\alpha$-particle has the largest mass, it possesses the shortest De Broglie wavelength. Hence, option 3 is the correct answer.

The radius of a nucleus of mass number $A$ is given by:

$$R = R_0 A^{1/3}$$

where $R_0 \approx 1.2 \times 10^{-15}\,\text{m}$ is a constant.

Taking the ratio of the two nuclear radii for $A_1 = 27$ and $A_2 = 8$:

$$\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{27}{8}\right)^{1/3} = \frac{3}{2}$$

The amount of radioactive substance remaining after time $t$ is given by:

$$N = N_0 \left(\frac{1}{2}\right)^n$$

where $n$ is the number of half-lives elapsed:

$$n = \frac{t}{t_{1/2}} = \frac{12\,\text{years}}{4\,\text{years}} = 3$$

Calculating the remaining mass:

$$N = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}$$

Matching components of a nuclear reactor with their functions:

• Uranium ($^{235}\text{U}$): Used as nuclear fission fuel (Matches III).
• Moderator (heavy water, graphite): Slows down fast fission neutrons to thermal energies to sustain controlled fission (Matches II).
• Control rods (cadmium, boron): Absorbs excess neutrons to regulate or halt the nuclear reaction rate (Matches I).
• Coolant (water, liquid sodium): Extracts and transfers thermal energy from reactor core to steam turbines to generate electricity (Matches IV).

Correct matching is (A)-(III), (B)-(II), (C)-(I), (D)-(IV). Hence, option 3 is the correct answer.