Geography High School

An electromagnetic wave A has a frequency(fA) 7 times higher than that of electromagnetic wave B(B). How do their wavelengths λA & λB da compare? 9. The spectrum of a blackbody has a peak wavelength of 7 x10^9 m. What is its temperature in kelvins?

Answers

Answer 1

The temperature of the blackbody is approximately 4.14 x 10^-13 kelvins.

The relationship between the frequency and wavelength of electromagnetic waves can be described by the equation λ = c/f, where λ is the wavelength, c is the speed of light, and f is the frequency.Given that electromagnetic wave A has a frequency (fA) that is 7 times higher than the frequency of electromagnetic wave B (fB), we can express this relationship as fA = 7fB.Since the speed of light (c) is constant, we can compare the wavelengths using the equation:λA = c/fA = c/(7fB) = (1/7)(c/fB) = (1/7)λB.Therefore, the wavelength of electromagnetic wave A (λA) is 1/7th of the wavelength of electromagnetic wave B (λB).
Given that the peak wavelength (λmax) is 7 x 10^9 m, we can rearrange the equation as:T = b/λmax = b/(7 x 10^9 m)
To calculate the temperature in kelvins, we need to know the value of the Wien's displacement constant (b), which is approximately 2.898 x 10^-3 m·K. Plugging this value into the equation:
T = (2.898 x 10^-3 m·K) / (7 x 10^9 m) ≈ 4.14 x 10^-13 K.
Therefore, the temperature of the blackbody is approximately 4.14 x 10^-13 kelvins.

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