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
[tex]\lambda_{max}[/tex] T = 2,898 10⁻³ [m K]
T = 2,898 10⁻³ / [tex]\lambda_{max}[/tex]
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
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
[tex]\texttt{ }[/tex]
Let's recall the Wien's Displacement Law as follows:
[tex]\boxed {\lambda_{max}\ T = 2.898 \times 10^{-3} \texttt{ m.K}}[/tex]
where:
λ_max = the wavelength of the maximum radiation energy ( m )
T = surface temperature of the star ( K )
Let us now tackle the problem!
[tex]\texttt{ }[/tex]
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:
[tex]T_r = ( 2.898 \times 10^{-3} ) \div \lambda_r[/tex]
[tex]T_r = ( 2.898 \times 10^{-3} ) \div ( 650 \times 10^{-9} )[/tex]
[tex]\boxed {T_r \approx 4500 \texttt{ K} }[/tex]
[tex]\texttt{ }[/tex]
[tex]T_y = ( 2.898 \times 10^{-3} ) \div \lambda_y[/tex]
[tex]T_y = ( 2.898 \times 10^{-3} ) \div ( 570 \times 10^{-9} )[/tex]
[tex]\boxed {T_y \approx 5100 \texttt{ K} }[/tex]
[tex]\texttt{ }[/tex]
[tex]T_b = ( 2.898 \times 10^{-3} ) \div \lambda_b[/tex]
[tex]T_b = ( 2.898 \times 10^{-3} ) \div ( 470 \times 10^{-9} )[/tex]
[tex]\boxed {T_b \approx 6200 \texttt{ K} }[/tex]
[tex]\texttt{ }[/tex]
[tex]T_w = ( 2.898 \times 10^{-3} ) \div \lambda_w[/tex]
[tex]T_w = ( 2.898 \times 10^{-3} ) \div ( 520 \times 10^{-9} )[/tex]
[tex]\boxed {T_w \approx 5600 \texttt{ K} }[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Mathematics
Chapter: Energy
