The wavelength of maximum intensity of radiation emitted by a star is 289.8 nm. The radiation intensity of the star is (Stefans constant = 5.67 × 10-8 W m-2 K-4, Wien's constant b = 2898 μ m K)

Question

The wavelength of maximum intensity of radiation emitted by a star is 289.8 nm. The radiation intensity of the star is (Stefans constant = 5.67 × 10-8 W m-2 K-4, Wien's constant b = 2898 μ m K) (A) 5.67 ×108 W m-2 (B) 5.67 ×1012 W m-2 (C) 10.67 × 107 W m-2 (D) 10.67 × 1014 W m-2

Tardigrade Admin · Accepted Answer

According to Wienâs displacement law, λmax T=b ∴ T=(b/λmax)=(2898 μ m K/289.8 nm) =(2898 ×10-6 m K/289.8 ×10-9 m) According to Stefan Boltzmann law, Radiation intensity, E=σ T4 E=(5.67 ×10-8 W m-2 K-4)(104 K)4 =5.67 ×108 W m-2

Q. The wavelength of maximum intensity of radiation emitted by a star is $289.8 nm$ . The radiation intensity of the star is
(Stefans constant $= 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$ , Wien's constant $b = 2898 μ m K$ )

$5.67 \times 1 0^{8} W m^{- 2}$

$5.67 \times 1 0^{12} W m^{- 2}$

$10.67 \times 1 0^{7} W m^{- 2}$

$10.67 \times 1 0^{14} W m^{- 2}$

Q. The wavelength of maximum intensity of radiation emitted by a star is 289.8nm. The radiation intensity of the star is (Stefans constant =5.67×10−8Wm−2K−4, Wien's constant b=2898μmK)

5.67×108Wm−2

5.67×1012Wm−2

10.67×107Wm−2

10.67×1014Wm−2

According to Wien’s displacement law, λmax​T=b ∴T=λmax​b​=289.8nm2898μmK​ =289.8×10−9m2898×10−6mK​ According to Stefan Boltzmann law, Radiation intensity, E=σT4 E=(5.67×10−8Wm−2K−4)(104K)4 =5.67×108Wm−2

Q. The wavelength of maximum intensity of radiation emitted by a star is $289.8 nm$ . The radiation intensity of the star is
(Stefans constant $= 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$ , Wien's constant $b = 2898 μ m K$ )

$5.67 \times 1 0^{8} W m^{- 2}$

$5.67 \times 1 0^{12} W m^{- 2}$

$10.67 \times 1 0^{7} W m^{- 2}$

$10.67 \times 1 0^{14} W m^{- 2}$