Two stars P and Q are observed at night. Star | vedclass

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(A) According to Wien's Displacement Law, the wavelength of maximum emission $(\lambda_{max})$ is inversely proportional to the absolute temperature $

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Temperature of $Q$ is higher than that of $P$

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(A) According to Wien's Displacement Law, the wavelength of maximum emission $(\lambda_{max})$ is inversely proportional to the absolute temperature $(T)$ of the star, given by $\lambda_{max} \propto 1/T$.Red light has a longer wavelength compared to white light (which contains all visible wavelengths, including shorter blue/violet wavelengths).Since star $P$ appears reddish, it emits light at a longer wavelength, implying a lower surface temperature.Since star $Q$ appears white, it emits light across the visible spectrum, including shorter wavelengths, implying a higher surface temperature.Therefore, the temperature of star $Q$ is higher than that of star $P$.

Two stars $P$ and $Q$ are observed at night. Star $P$ appears reddish while star $Q$ is white. From this we conclude:

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