A small object is placed at the center of a large evacuated hollow spherical container. Assume that the container is maintained at 0 K. At time $t = 0,$ the temperature of the object is 200 K. The temperature of the object becomes 100 K at $t = t_{1}$ and 50 K at $t = t_{2}.$ Assume the object and the container to be ideal black bodies. The heat capacity of the object does not depend on temperature. The ratio $(t_{2}/t_{1})$ is ___________.

Step 1: Express the net rate of heat loss.

By the Stefan-Boltzmann law, the net power radiated by the black body to the container at $0\,\text{K}$ is:

$$P_{\text{net}} = \sigma A T^4$$

Step 2: Formulate the differential equation.

The rate of change of the object's thermal energy is related to its heat capacity $C$ and temperature $T$:

$$C \frac{dT}{dt} = -\sigma A T^4$$

Step 3: Separate variables and integrate.

Rearrange the terms to separate the variables $T$ and $t$:

$$\frac{dT}{T^4} = -\frac{\sigma A}{C} dt$$

Integrate both sides from the initial state at $t = 0$ where $T = T_0$:

$$\int_{T_0}^{T} \frac{dT}{T^4} = -\frac{\sigma A}{C} \int_{0}^{t} dt$$ $$-\frac{1}{3}\left(\frac{1}{T^3} - \frac{1}{T_0^3}\right) = -\frac{\sigma A}{C} t$$

Step 4: Solve for time $t$.

Express the time $t$ as a function of temperature $T$:

$$t = \frac{C}{3\sigma A} \left(\frac{1}{T^3} - \frac{1}{T_0^3}\right)$$

Step 5: Write expressions for $t_1$ and $t_2$.

Substitute the given temperatures $T_0 = 200\,\text{K}$, $T = 100\,\text{K}$ for $t_1$, and $T = 50\,\text{K}$ for $t_2$:

$$t_1 = \frac{C}{3\sigma A} \left(\frac{1}{100^3} - \frac{1}{200^3}\right)$$ $$t_2 = \frac{C}{3\sigma A} \left(\frac{1}{50^3} - \frac{1}{200^3}\right)$$

Step 6: Calculate the ratio of the times.

Divide $t_2$ by $t_1$ to find the ratio:

$$\frac{t_2}{t_1} = \frac{\dfrac{1}{50^3} - \dfrac{1}{200^3}}{\dfrac{1}{100^3} - \dfrac{1}{200^3}}$$

Multiply the numerator and denominator by $200^3$:

$$\frac{t_2}{t_1} = \frac{\left(\dfrac{200}{50}\right)^3 - 1}{\left(\dfrac{200}{100}\right)^3 - 1}$$ $$\frac{t_2}{t_1} = \frac{4^3 - 1}{2^3 - 1}$$ $$\frac{t_2}{t_1} = \frac{63}{7}$$ $$\frac{t_2}{t_1} = 9$$

Answer: $9$