Question

The maximum wavelength of radiation emitted by a star is 289.8 nm. Then intensity of radiation for the star is ______. 
(Given: Stefan's constant= 5.67 x 10^-8 Wm^-2 K^-4, Wien's constant, b = 2898 µ mK).

पर्याय

• 5.67 x 10^-12 Wm^-2

• 10.67 x 10¹⁴ Wm^-2

• 5.67 x 10⁸ Wm^-2

• 10.67 x 10⁷ Wm^-2

Answer 1

The maximum wavelength of radiation emitted by a star is 289.8 nm. Then intensity of radiation for the star is 5.67 x 10⁸ Wm^-2.

Explanation:

Given, maximum wavelength,

`lambda_"m" = 289.8 "nm" = 289.8 xx 10^-9 "m"`

`= 2898 xx 10^-10`m

Stefan's constant, `sigma = 5.67 xx 10^-8 "Wm"^-2 "K"^-4`

Wien's constant, b = 2898 µmK = 2898 x 10^-6 mK

According to Wien's displacement law, the maximum wavelength is given by

`lambda_"m" = "b"/"T" => "T" = "b"/lambda_"m"`    ...(i)

Substituting given values in Eq. (i), we get

T = `(2898 xx 10^-6)/(2898 xx 10^-10) = 10^4`K     ...(ii)

According to Stefan's law, the energy radiated from a source is given by

`"E" = sigma "Ae T"^4`     ...(iii)

where, A = area of source

e = emissivity (value between 0 to 1)

The intensity of radiations emitted is equal to energy radiated from a given surface area, i.e.,

I = `"E"/"A" = sigma "e" "T"^4`    ..[from Eq. (iii)]

Ase is very small, so

I = σT⁴       .....(iv)

Substituting the value of T from Eq. (ii) in Eq. (iv), we get

I = `sigma (10^4)^4 = 5.67 xx 10^-6 xx 10^16   ...[because sigma = 5.67 xx 10^-8  ("given")]`

`= 5.67 xx 10^8  "Wm"^-2`