Question
Use the formula $\lambda_{ m } T =0.29 cm K$ to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?
✓
Answer
A body at a particular temperature produces a continuous spectrum of wavelengths. In the case of a black body, the wavelength corresponding to the maximum intensity of radiation is given according to Planck's law. It can be given by the relation,
$\lambda_m=\frac{0.29}{T} cmK$
Where,
$\lambda_{ m }=$ maximum wavelength
T = temperature
Thus, the temperature for different wavelengths can be obtained as:
For $\lambda_{ m }=10^{-4} cm ; T=\frac{0.29}{10^{-4}}=2900^{\circ} k$
For $\lambda_{ m }=5 \times 10^{-5} cm ; T=\frac{0.29}{5 \times 10^{-5}}=5800^{\circ} k$
For $\lambda_{ m }=10^{-6} cm ; T=\frac{0.29}{10^{-6}}=290000^{\circ} k$ and so on.
The numbers obtained tell us that temperature ranges are required for obtaining radiations in different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding temperature increases.
$\lambda_m=\frac{0.29}{T} cmK$
Where,
$\lambda_{ m }=$ maximum wavelength
T = temperature
Thus, the temperature for different wavelengths can be obtained as:
For $\lambda_{ m }=10^{-4} cm ; T=\frac{0.29}{10^{-4}}=2900^{\circ} k$
For $\lambda_{ m }=5 \times 10^{-5} cm ; T=\frac{0.29}{5 \times 10^{-5}}=5800^{\circ} k$
For $\lambda_{ m }=10^{-6} cm ; T=\frac{0.29}{10^{-6}}=290000^{\circ} k$ and so on.
The numbers obtained tell us that temperature ranges are required for obtaining radiations in different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding temperature increases.
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