Our Sun emits most of its radiation at a wavelength of 550 nm. If a star were 4.00 times hotter than our Sun, it would emit most of its radiation at a wavelength of .

88.0015184nm

Explanation:

By wein's law

λT= 2.898x10^{-3}kelvin meter

where;

λ = wavelenght

T= Temperature (in kelvin)

2.898x10^{-3}mk = Wein's constant

so

λ x T(sun)= 2.898x10^{-3}

550 x 10\small ^{-9}x T (sun) =2.898x10^{-3}

T (sun) = 2.898x10^{-3}/ 550 x 10\small ^{-9}

T (sun) = 5269.09090909 kelvin

T (sun) = 5.269x10^{3}kelvin

Now, Given T( star )= 4 x T (sun)

so,

λ (star) x T( star ) = 2.898x10^{-3}kelvin meter

λ(star) x 4 x T (sun) = 2.898x10^{-3}kelvin meter

λ (star) x 4 x 5.269x10^{3}kelvin = 2.898x10^{-3}kelvin meter

λ(star) = ( 2.898x10^{-3}kelvin meter)/ (4 x 5.269x10^{3}kelvin)

λ (star) = 8.80015184 x 10^{-7}meter

λ (star) = 88.0015184 x 10^{-9}meter

λ (star) = 88.0015184 nm