Explanation Given data: Wavelength of Sirius is 290 nm. Wavelength of sun is 500 nm. sirius diameter = 1.7 × sun diameter . Formula used: Consider the expression of temperature. T = 2.9 × 10 6 nm ⋅ K λ (1) Here, λ is the wavelength. Stefan-Boltzmann law is given by, Δ Q Δ t = e σ A T 4 (2) Here, Δ Q is the energy, Δ t is the time interval, e is the emissivity of the surfaces, σ is Stefan–Boltzmann constant, and A is the area. Explanation: For Sirius temperature, Substitute 290 nm for λ in equation (1), T sirius = 2.9 × 10 6 nm ⋅ K 290 nm = 1 × 10 4 K For Sun temperature, Substitute 500 nm for λ in equation (1), T sun = 2.9 × 10 6 nm ⋅ K 500 nm = 5.8 × 10 3 K For Sirius energy is given by equation (2), Δ Q sirius Δ t = e σ A sirius T sirius 4 (3) For Sun energy is given by equation (2), Δ Q sun Δ t = e σ A sun T sun 4 (4) On dividing equation (3) and equation (4), Δ Q sirius Δ t Δ Q sun Δ t = e σ A sirius T sirius 4 e σ A sun T sun 4 Δ Q sirius Δ Q sun = A sirius T sirius 4 A sun T sun 4 As Sirius is a circle so area is, A sirius = π × ( R sirius ) 2 Here, R sirius is the radius

College Physics: A Strategic Approach (3rd Edition)

3rd Edition

ISBN: 9780321879721

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

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Nature of light

Light is transverse and electromagnetic radiation that exhibits the properties of a wave as well as a particle. For this reason, light is said to have a “dual nature.”

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Chapter 25, Problem 54P

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The factor by which the energy emitted by Sirius exceed that of sun.

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Chapter 25, Problem 53P

Chapter 25, Problem 55P

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College Physics: A Strategic Approach (3rd Edition)

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The factor by which the energy emitted by Sirius exceed that of sun.

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