Question

Our Sun emits most of its radiation at a wavelength of 550nm. If a star were 5.00 times hotter than our Sun, it would emit most of its radiation at a wavelength of nm 1st attempt Feedback Q See Hint Our Sun emits most of its radiation at a wavelength of 550 nm. If a star were 5.00 times hotter than our Sun, it would emit most of its radiation at a wavelength of 275 nm

Answer 1

Answer:

$\lambda_{star} = 1.100\times 10^{-7} meter$

Explanation:

By wein's law

$\lambda\times T= 2.898\times 10^{-3} kelvin meter$

where;

λ = wavelenght

T= Temperature (in kelvin)

$2.898\times 10^{-3} mk$ = Wein's constant

so

$\lambda\times T(sun) = 2.898\times 10^{-3}$

$550\times 10^{-9}\times T (sun) = 2.898\times 10^{-3}$

$T (sun) = \frac{2.898\times 10^{-3}}{ 550\times 10 ^{-9}}$

T (sun) = 5269.09 kelvin

from data given we have

$T( star ) = 5\times T (sun)$

so,

$\lambda_{star}\times T(star) = 2.898\times 10^{-3}$

$\lambda_{star}\times 5\times T(sun) = 2.898\times 10^{-3}kelvin meter$

$\lambda_{star}\times 5\times 5269.09 = 2.898\times 10^{-3}kelvin meter$

$\lambda_{star} = \frac{(2.898\times 10^{-3}kelvin meter)}{(5 \times 5.269 \times 10^{3}kelvin)}$

$\lambda_{star} = 1.100\times 10^{-7} meter$