A blue supergiant star has surface temperature 27 kK and has total power output 100000 times that of the Sun.

a. Where is the Wien peak of this star's blackbody curve?​ λmax =

b. Given the Sun's 6.96×108m radius, what's the radius of the supergiant? r =

Respuesta :

Answer:

Wien peak ( λmax ) is 107.40 nm

radius of super giant is 1.086 ×[tex]10^{10}[/tex] m

Explanation:

given data

temperature 27 kK

power = 100000 times of Sun

Sun radius =  6.96 × 10^8 m

to find out

Wien peak ( λmax ) and radius of supergiant (r)

solution

we will apply here first wien law to find Wien peak  that is

λmax = b / t

λmax = 2.9 × [tex]10^{-3}[/tex] / 27000 = 1.0740 × [tex]10^{-7}[/tex]

so Wien peak ( λmax ) is 107.40 nm

and

now we apply steafay law that is

P = σ × A × [tex]T^{4}[/tex]   .........................1

and we know total power output 100000 time of Sun

so we say  

4πr²s[tex]T^{4}[/tex]  = 100000 × 4πR²s[tex]Ts^{4}[/tex]

r² = 100000  × R²[tex]Ts^{4}[/tex] / [tex]T^{4}[/tex]

put here value

r² = 100000  × (6.96×[tex]6000^{8}[/tex] )² × [tex]6000^{4}[/tex] / [tex]27000^{4}[/tex]

r² = 1.18132 ×[tex]10^{20}[/tex]

r = 1.086 ×[tex]10^{10}[/tex] m

so radius of super giant is 1.086 ×[tex]10^{10}[/tex] m

ACCESS MORE