(c) A black body at temperature 4980 K emits radiation of wavelength 4000 Å with maximum energy. Calculate the temperature at which it will emit a wavelength of \(1.45 \times 10^{-5} \, \text{cm}\) with maximum energy.

Another Explanation (5): Using Wien's displacement law, \( \lambda_{max} T = b \), where \(b\) is Wien's displacement constant (\(2.898 \times 10^{-3} \, m \cdot K\)). Given \( \lambda_{max1} = 4000 \, Å = 4000 \times 10^{-10} \, m\) and \(T_1 = 4980 \, K\). We have \(b = \lambda_{max1} T_1 = (4000 \times 10^{-10} \, m)(4980 \, K) = 1.992 \times 10^{-3} \, m \cdot K\) For the second wavelength, \( \lambda_{max2} = 1.45 \times 10^{-5} \, cm = 1.45 \times 10^{-7} \, m\). Then, \(T_2 = \frac{b}{\lambda_{max2}} = \frac{1.992 \times 10^{-3} \, m \cdot K}{1.45 \times 10^{-7} \, m} \approx 13740 \, K\) Therefore, the temperature is approximately 13740 K.