Physics · Thermal Properties Of Matter · NEET
More power. A shorter peak wavelength means a higher temperature (Wien: lambda_max is proportional to 1/T). Higher temperature means much higher power (Stefan: P is proportional to T^4). So a shorter peak wavelength always goes with a hotter, brighter body radiating more total power.
Use temperature as the bridge. Step 1: from the wavelength change, find the temperature ratio using T'/T = lambda_max / lambda_max'. Note it is inverted because T and lambda_max are inversely related. Step 2: put that temperature ratio into P'/P = (T'/T)^4. That single number is the power ratio.
Because lambda_max is proportional to 1/T, not to T. So if the peak wavelength becomes (3/4) of the old value, the temperature becomes (4/3) of the old value. You must flip the wavelength ratio to get the temperature ratio, then raise to the 4th power. Forgetting to flip is the number one mistake.
T'/T = lambda_0 / ((3/4)lambda_0) = 4/3. Then P'/P = (4/3)^4 = 256/81, which is about 3.16. So the new power is 256/81 times the old power. This is exactly the NEET 2018 answer.
The clean P is proportional to T^4 form is for a black body. For a real (grey) body, P = e sigma A T^4 with emissivity e. As long as the size (area) and emissivity stay the same, the ratio P'/P still equals (T'/T)^4, so the method is unchanged. Only the size or emissivity changing would add extra factors.
The power radiated by a black body is P and it radiates maximum energy at wavelength lambda_0. If the temperature is changed so that it now radiates maximum energy at wavelength (3/4)lambda_0, the power radiated becomes nP. The value of n is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Get the temperature ratio by inverting the peak wavelength ratio (T'/T = lambda_max/lambda_max'), then raise it to the 4th power for the power ratio (P'/P = (T'/T)^4).
Doubling lambda_max means temperature halves (T'/T = 1/2). Power becomes (1/2)^4 = 1/16 of the old value. The body is cooler and much dimmer.
Wien's displacement law (lambda_max is proportional to 1/T, constant b = 2.9 x 10^-3 m K) and the Stefan-Boltzmann law (P is proportional to T^4).
No. In ratio problems the constant cancels. You only need the ratio of the two peak wavelengths. The constant is needed only when a numerical temperature value is asked.
It is the total power radiated over all wavelengths, given by Stefan's law P is proportional to T^4. Do not confuse it with the energy emitted at one single wavelength.
