A black body at 200 K is found to exit maximum energy at a wavelength of 14 - m- When its temperature is raised to 1000 K- the wavelength at which maximum energy is emitted isA- 14 - mB- 70 - FC- 2-8 - mD- 2-04 mm

The correct option is C 2.8 μm T_1λ _m_1=T_2λ _m_2⇒λ _m_2=T_1 λ _m_1/T_2=14 × 200/1000=2.8 μ m ...

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Standard XII

Physics

Blackbody

A black body ...

Question

A black body at 200 K is found to exit maximum energy at a wavelength of 14 $μ$ m. When its temperature is raised to 1000K, the wavelength at which maximum energy is emitted is

14 μm

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70 μF

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2.8 μm

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2.04 mm

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Solution

The correct option is C
2.8 μm

$T_{1} λ_{m_{1}} = T_{2} λ_{m_{2}} \Rightarrow λ_{m_{2}} = \frac{T_{1} λ_{m_{1}}}{T_{2}} = \frac{14 \times 200}{1000} = 2.8 μ m$

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A black body at 200 K is found to exit maximum energy at a wavelength of 14 μ m. When its temperature is raised to 1000K, the wavelength at which maximum energy is emitted is

The correct option is C 2.8 μm T1λm1=T2λm2⇒λm2=T1λm1T2=14×2001000=2.8μm

A black body at 200 K is found to exit maximum energy at a wavelength of 14 $μ$ m. When its temperature is raised to 1000K, the wavelength at which maximum energy is emitted is

The correct option is C
2.8 μm

$T_{1} λ_{m_{1}} = T_{2} λ_{m_{2}} \Rightarrow λ_{m_{2}} = \frac{T_{1} λ_{m_{1}}}{T_{2}} = \frac{14 \times 200}{1000} = 2.8 μ m$