f) If a blackbody at temperature 6174 K emits 4700 Å with maximum energy, calculate the temperature at which it will emit a wavelength of \(1.4 \times 10^{-3}\) m with maximum energy.
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JU-PHY2nd YearFinalThermal PhysicsRadiationStefan-Boltzmann law; Wien's displacement law. (Topic Practice)JU-PHY - ⚡ অনলাইন প্রশ্নব্যাংক দেখুন 💥
Another Explanation (5): Using Wien's Displacement Law, λmaxT = constant.
Let λ1 = 4700 Å = 4700 × 10-10 m and T1 = 6174 K.
Let λ2 = 1.4 × 10-3 m and T2 be the unknown temperature.
Then λ1T1 = λ2T2
Therefore, T2 = (λ1T1) / λ2 = (4700 × 10-10 m × 6174 K) / (1.4 × 10-3 m) ≈ 2.08 K
Related Questions (Any University/Year)
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