Question

What is the surface temperature of a distant star having a peak wavelength of 475 nm?

Answer 1

Answer:

Explanation:

Given that,

The peak wavelength is

λp = 475nm

Then, we want to find the temperature,

From Wein's displacement law,

When the maximum is evaluated from the Planck radiation formula, the product of the peak wavelength and the temperature is found to be a constant (k = 2.898 ×10^-3 mK)

So, applying this we have

λp•T = 2.898 ×10^-3

T = 2.898 × 10^-3 / λp

T = 2.898 × 10^-3 / 475 × 10^-9

T = 6101.05 K

Answer 2

The surface temperature of the distant star having the given peak wavelength is 6.1 × 10³K.

Given the data in the question;'

Peak wavelength; $\lambda _p = 475nm = 4.75 *10^{-7}m$

To determine the surface temperature the star, we use the expression for peak wavelength by Wien's Displacement Law:

$\lambda _p = \frac{0.2898 * 10^{-2}m.K}{T}$

Where T is the surface temperature

We substitute our value into the formula and solve for for "T"

$4.75*10^{-7}m = \frac{0.2898 * 10^{-2}m.K}{T}\\\\T = \frac{0.2898 * 10^{-2}m.K}{4.75*10^{-7}m}\\\\T = 6. 1 * 10^3K$

Therefore, the surface temperature of the distant star having the given peak wavelength is 6.1 × 10³K.

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