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

$14 μm$

$70 μF$

$2.8 μm$

$2.8 nm$

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Explanation

(c) $λ_{m 1} T_{1} = λ_{m 2} T_{2} \Rightarrow λ_{m 2} = \frac{λ_{m 1} T_{1}}{T_{2}} = \frac{14 \times 200}{1000} = 2.8 μm$

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If the temperature of the sun becomes twice its present temperature, then

Radiated energy would be predominantly in infrared

Radiated energy would be predominantly in ultraviolet

Radiated energy would be predominantly in X-ray region

Radiated energy would become twice the present radiated energy

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Explanation

(b) The wavelength corresponding to maximum emission of radiation from the sun is $λ_{\max} = 4753 \overset{0}{A}$ (close to the wavelength of violet colour of visible region). Hence if temperature is doubled $λ_{m}$ is decreased $(λ_{m} \propto \frac{1}{T})$ i.e. mostly ultraviolet radiations emits.

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The maximum energy in the thermal radiation from a hot source occurs at a wavelength of $11 \times 10^{- 5}$ cm. According to Wein's law, the temperature of the source (on Kelvin scale) will be n times the temperature of another source (on Kelvin scale) for which the wavelength at maximum energy is $5.5 \times 10^{- 5}$ cm. The value n is

$\frac{1}{2}$

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Explanation

(c) $\frac{T_{1}}{T_{2}} = \frac{λ_{m 2}}{λ_{m 1}} = \frac{5.5 \times 10^{5}}{11 \times 10^{5}} = \frac{1}{2} \Rightarrow n = \frac{1}{2}$

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How is the temperature of stars determined by ?

Stefan’s law

Wein’s displacement law

Kirchhoff’s law

Ohm’s law

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Explanation

(b) The temperature of stars is determined by using the wavelength of light coming from them in Wan's displacement law.

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On increasing the temperature of a substance gradually, which of the following colours will be noticed by you ?

White

Yellow

Green

Red

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Explanation

(a)

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A black body has maximum wavelength $λ_{m}$ at temperature 2000 K. Its corresponding wavelength at temperature 3000 K will be

$\frac{3}{2} λ_{m}$

$\frac{2}{3} λ_{m}$

$\frac{4}{9} λ_{m}$

$\frac{9}{4} λ_{m}$

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Explanation

(b) $λ_{m_{2}} = \frac{T_{1}}{T_{2}} \times λ_{m_{1}} = \frac{2000}{3000} \times λ_{m_{1}} = \frac{2}{3} λ_{m_{1}} = \frac{2}{3} λ_{m}$

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A black body at a temperature of 1640 K has the wavelength corresponding to maximum emission equal to 1.75 $μm$ . Assuming the moon to be a perfectly black body, the temperature of the moon, if the wavelength corresponding to maximum emission is 14.35 $μ$ m is

100 K

150 K

200 K

250 K

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Explanation

(c) $\frac{T_{2}}{T_{1}} = \frac{λ_{m_{1}}}{λ_{m_{2}}} = \frac{1.75}{14.35} \Rightarrow T_{2} = \frac{1.75}{14.35} \times 1640 = 200 K$

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A particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^{4} K$ . The wavelength in nanometers at which its radiation becomes maximum is -
(b = 0.0029 mK)

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Explanation

(b) According to Wein’s displacement law

$λ_{m} T = b or λ_{m} = \frac{b}{T} = \frac{0.0029}{5 \times 10^{4}} = 58 \times 10^{- 9} m = 58 nm$

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The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and that emitted by the north star has the maximum value at 350 nm. If these stars behave like black bodies, then the ratio of the surface temperature of the sun and north star is

1.46

0.69

1.21

0.83

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Explanation

(b) $\frac{T_{S}}{T_{N}} = \frac{{(λ_{N})}_{\max}}{{(λ_{S})}_{\max}} = \frac{350}{510} = 0.69$

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The amount of radiation emitted by a perfectly black body is proportional to

Temperature on ideal gas scale

Fourth root of temperature on ideal gas scale

Fourth power of temperature on ideal gas scale

Source of temperature on ideal gas scale

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Explanation

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