Choice A is the correct answer. Solve for $T$ when $R(T) = 5$.

1. Equation: $5 = 50e^{-0.1T}$.
2. Divide: $0.1 = e^{-0.1T}$.
3. Natural log: $\ln(0.1) = -0.1T$.
4. Solve: $T = \frac{\ln(0.1)}{-0.1} = \frac{-2.303}{-0.1} \approx 23.03 \approx 23$ K.

💡 Strategic Tip:$\ln(0.1) = -\ln(10) \approx -2.303$.

Choice B is incorrect because this is too high. Choice C is incorrect because this is too low. Choice D is incorrect because this is way too high.