To determine the peak wavelength of the radiation emitted by Epsilon Indi, we can use Wien's Displacement Law. This law states that the wavelength at which the emission of a black body spectrum is at its maximum, λmax, is inversely proportional to the temperature of the black body, T. The formula is:

λmax = b / T

where b is Wien's constant, which is approximately 2.898 x 10^-3 m.K.

Given that the surface temperature of Epsilon Indi, T, is 4600 K, we can substitute this into the formula to find λmax:

λmax = (2.898 x 10^-3 m.K) / 4600 K

Solving this gives:

λmax ≈ 6.30 x 10^-7 m, or 630 nm.

So, the peak wavelength of the radiation emitted by Epsilon Indi is approximately 630 nm.

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To determine the peak wavelength of the radiation emitted by Epsilon Indi, we can use Wien's Displacement Law. This law states that the wavelength at which the emission of a black body spectrum is at its maximum, λmax, is inversely proportional to the temperature of the black body, T. The formula is:

λmax = b / T

where b is Wien's constant, which is approximately 2.898 x 10^-3 m.K.

Given that the surface temperature of Epsilon Indi, T, is 4600 K, we can substitute this into the formula to find λmax:

λmax = (2.898 x 10^-3 m.K) / 4600 K

Solving this gives:

λmax ≈ 6.30 x 10^-7 m, or 630 nm.

So, the peak wavelength of the radiation emitted by Epsilon Indi is approximately 630 nm.

Knowee AI · Accepted Answer

To determine the peak wavelength of the radiation emitted by Epsilon Indi, we can use Wien's Displacement Law. This law states that the wavelength at which the emission of a black body spectrum is at its maximum, λmax, is inversely proportional to the temperature of the black body, T. The formula is:

λmax = b / T

where b is Wien's constant, which is approximately 2.898 x 10^-3 m.K.

Given that the surface temperature of Epsilon Indi, T, is 4600 K, we can substitute this into the formula to find λmax:

λmax = (2.898 x 10^-3 m.K) / 4600 K

Solving this gives:

λmax ≈ 6.30 x 10^-7 m, or 630 nm.

So, the peak wavelength of the radiation emitted by Epsilon Indi is approximately 630 nm.

The surface temperature of the star Epsilon Indi is 4600 K.(a) (i) Determine the peak wavelength of the radiation emitted by Epsilon Indi

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