Question

The colors of the stars in the sky range from red to blue. Assuming that the color indicates the frequency at which the star radiates the maximum amount of electromagnetic energy, estimate the surface temperature of red, yellow, white, and blue stars.
A) Estimate the surface temperature of red star. Assume that red color corresponds to wavelengths of approximately 650 nm.B) Estimate the surface temperature of yellow star. Assume that yellow color corresponds to wavelengths of approximately 570 nm.C) Estimate the surface temperature of blue star. Assume that blue color corresponds to wavelengths of approximately 470 nmD) Estimate the surface temperature of white star. Since green is in the middle of the visible spectrum, choose 520 nm wavelength for white star.

Answer 1

A ) the surface temperature of red star is about 4500 K

B ) the surface temperature of yellow star is about 5100 K

C ) the surface temperature of blue star is about 6200 K

D ) the surface temperature of white star is about 5600 K

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Further explanation

Let's recall the Wien's Displacement Law as follows:

$\boxed {\lambda_{max}\ T = 2.898 \times 10^{-3} \texttt{ m.K}}$

where:

λ_max = the wavelength of the maximum radiation energy ( m )

T = surface temperature of the star ( K )

Let us now tackle the problem!

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Given:

wavelength of red light = λ_r = 650 nm = 650 × 10⁻⁹ m

wavelength of yellow light =  λ_y = 570 nm = 570 × 10⁻⁹ m

wavelength of blue light =  λ_b = 470 nm = 470 × 10⁻⁹ m

wavelength of white light =  λ_w = 520 nm = 520 × 10⁻⁹ m

Asked:

A ) the surface temperature of red star = T_r = ?

B ) the surface temperature of yellow star = T_y = ?

C ) the surface temperature of blue star = T_b = ?

D ) the surface temperature of white star = T_w = ?

Solution:

Part A) :

$T_r = ( 2.898 \times 10^{-3} ) \div \lambda_r$

$T_r = ( 2.898 \times 10^{-3} ) \div ( 650 \times 10^{-9} )$

$\boxed {T_r \approx 4500 \texttt{ K} }$

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Part B) :

$T_y = ( 2.898 \times 10^{-3} ) \div \lambda_y$

$T_y = ( 2.898 \times 10^{-3} ) \div ( 570 \times 10^{-9} )$

$\boxed {T_y \approx 5100 \texttt{ K} }$

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Part C) :

$T_b = ( 2.898 \times 10^{-3} ) \div \lambda_b$

$T_b = ( 2.898 \times 10^{-3} ) \div ( 470 \times 10^{-9} )$

$\boxed {T_b \approx 6200 \texttt{ K} }$

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Part D) :

$T_w = ( 2.898 \times 10^{-3} ) \div \lambda_w$

$T_w = ( 2.898 \times 10^{-3} ) \div ( 520 \times 10^{-9} )$

$\boxed {T_w \approx 5600 \texttt{ K} }$

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Answer details

Grade: High School

Subject: Mathematics

Chapter: Energy

Answer 2

Answer:

a) 4458K b) 5048K, c) 6166K, d) 5573K

Explanation:

The temperature of the stars and many very hot objects can be estimated using the Wien displacement law

    $\lambda_{max}$ T = 2,898 10⁻³ [m K]

    T = 2,898 10⁻³ / $\lambda_{max}$

a) indicate that the wavelength is

    Lam = 650 nm (1 m / 109 nm) = 650 10⁻⁹ m

    Lam = 6.50 10⁻⁷ m

    T = 2,898 10⁻³ / 6.50 10⁻⁷

    T = 4,458 10³ K

    T = 4458K

b) lam = 570 nm = 5.70 10⁻⁷ m

    T = 2,898 10⁻³ / 5.70 10⁻⁷

    T = 5084K

c) lam = 470 nm = 4.70 10⁻⁷ m

    T = 2,898 10⁻³ / 4.7 10⁻⁷

    T = 6166K

d) lam = 520 nm = 5.20 10⁻⁷ m

    T = 2,898 10⁻³ / 5.20 10⁻⁷

    T = 5573K