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NCERT Class XII Chapter
Electromagnetic Waves
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Question : 13 of 15
Marks: +1, -0
Use the formula λmT\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?
Solution:  
A black body at very high temperature produce a continuous spectrum. Using Wien’s displacement law we can calculate the wavelength corresponding to maximum intensity of radiation emitted Required absolute temperature
λmT\lambda_m T = 0.29 cm K or T = 0.29×10−2λm\frac{0.29 \times 10^{-2}}{\lambda_m} K
Using above formula, the temperature of black body required for various wavelength is calculated.
Type of radiation Wavelength of radiation Temperature required
1. Radio waves 1 m 29 × 10−410^{-4} K
2. Micro waves 3 × 10−310^{-3} m 1 K
3. Infrared waves 3 × 10−510^{-5} m 10210^{2} K
4. Visible waves 5 × 10−710^{-7} m 6 × 10310^{3} K
5. Ultraviolet rays 3 × 10−710^{-7} m 10410^{4} K
6. X-rays 10−1010^{-10} m 29 × 10610^{6} K
7. γ rays 10−1210^{-12} m 29 × 10810^{8} K
To produce electromagnetic radiations of different wavelength, we need temperature ranges. To produce visible radiation of l = 5 × 10−710^{-7} m, we need to have source at temperature of 6000 K. A source at lower temperature will produce this wavelength but not with maximum intensity
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