The wavelength of maximum emitted energy of a body at 700 K is 4.08 $μ m$ If the temperature of the body is raised to 1400 K, the wavelength of maximum emitted energy will be

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Solution

The correct option is D

$λ_{m 1} T_{1} = λ_{m 1} T_{2} \Rightarrow λ_{m 2} = \frac{λ_{m 1} T_{1}}{T_{2}} = 4.08 \times \frac{700}{1400} = 2.04 m$

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The wavelength of maximum emitted energy of a body at 700 K is 4.08μm If the temperature of the body is raised to 1400 K, the wavelength of maximum emitted energy will be

The correct option is D λm1T1=λm1T2⇒λm2=λm1T1T2=4.08×7001400=2.04 m

The wavelength of maximum emitted energy of a body at 700 K is 4.08 $μ m$ If the temperature of the body is raised to 1400 K, the wavelength of maximum emitted energy will be

The correct option is D

$λ_{m 1} T_{1} = λ_{m 1} T_{2} \Rightarrow λ_{m 2} = \frac{λ_{m 1} T_{1}}{T_{2}} = 4.08 \times \frac{700}{1400} = 2.04 m$