Two spherical stars $A$ and $B$ emit blackbody radiation. The radius of $A$ is 400 times that of $B$ and $A$ emits $10^{4}$ times the power emitted from $B$. The ratio $\left(\dfrac{\lambda_A}{\lambda_B}\right)$ of their wavelengths $\lambda_A$ and $\lambda_B$ at which the peaks occur in their respective radiation curves is

Step 1: Identify the given parameters.

The relationship between the radii and powers of the two spherical stars $A$ and $B$ is given.

$$R_A = 400 R_B$$ $$P_A = 10^4 P_B$$

Step 2: Apply the Stefan-Boltzmann Law.

The total power $P$ emitted by a spherical blackbody of radius $R$ at temperature $T$ is given by the Stefan-Boltzmann Law.

$$P = 4\pi R^2 \sigma T^4$$

Step 3: Determine the ratio of temperatures.

Express the ratio of the powers of the two stars.

$$\frac{P_A}{P_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_A}{T_B}\right)^4$$

Substitute the given values into the ratio.

$$10^4 = (400)^2 \left(\frac{T_A}{T_B}\right)^4$$ $$10^4 = 1.6 \times 10^5 \left(\frac{T_A}{T_B}\right)^4$$

Solve for the ratio of temperatures.

$$\left(\frac{T_A}{T_B}\right)^4 = \frac{10^4}{1.6 \times 10^5}$$ $$\left(\frac{T_A}{T_B}\right)^4 = \frac{1}{16}$$ $$\frac{T_A}{T_B} = \frac{1}{2}$$

Step 4: Apply Wien's Displacement Law.

Wien's Displacement Law relates the peak wavelength $\lambda$ to the temperature $T$.

$$\lambda T = b$$ $$\lambda \propto \frac{1}{T}$$

Step 5: Calculate the ratio of the peak wavelengths.

The ratio of the peak wavelengths is inversely proportional to the ratio of their temperatures.

$$\frac{\lambda_A}{\lambda_B} = \frac{T_B}{T_A}$$

Substitute the temperature ratio.

$$\frac{\lambda_A}{\lambda_B} = 2$$

Answer: 2