In the following nuclear reaction,
$$D \xrightarrow{\alpha} D_1 \xrightarrow{\beta} D_2 \xrightarrow{\alpha} D_3 \xrightarrow{\gamma} D_4$$
Mass number of $$D$$ is 182 and atomic number is 74. Mass number and atomic number of $$D_4$$ respectively will be

We need to track the mass number (A) and atomic number (Z) through a series of nuclear decays. For nucleus D, A = 182 and Z = 74.

In alpha decay (D → D₁), mass number decreases by 4 and atomic number decreases by 2.

$$D_1: A = 182 - 4 = 178, \quad Z = 74 - 2 = 72$$

In beta (β⁻) decay (D₁ → D₂), mass number stays the same and atomic number increases by 1.

$$D_2: A = 178, \quad Z = 72 + 1 = 73$$

Another alpha decay (D₂ → D₃) gives:

$$D_3: A = 178 - 4 = 174, \quad Z = 73 - 2 = 71$$

In gamma decay (D₃ → D₄), neither mass number nor atomic number changes (only energy is released).

$$D_4: A = 174, \quad Z = 71$$

The mass number and atomic number of D₄ are 174 and 71 respectively. The answer is Option A: 174 and 71.