Numerical problems of Wien displacement law

1. Numerical problems on Wien displacement law 1 JAYAM CHEMISTRY ADDA

2. What is the temperature of an astronomical object present in a star galaxy Problem-1 with a wavelength of about 1.42 x 10-5cm? Answer The wavelength of the astronomical object = 1.42 x 10-5cm According to Wien displacement law, we have; λmx T = b b T = λm Value of b=0.289 cm K 0.289 cm K 1.42 × 10−5cm T = 0.20352 x 105K T = T = 20352 K 2 JAYAM CHEMISTRY ADDA

3. The multiplicative product of λm and the body's temperature is one- Problem-3 fourth of a constant quantity 'u' having a value of 1.156 cm Kelvin. The body's temperature is 18000 Kelvin, then find λm. According to the question, we have; Answer λmx T = u/4 Temperature of the body = 18000 Kelvin λ?× 18000 ?????? =1.156 ?? ?????? 4 1.156 ?? ?????? 4 × 18000 ?????? λ?= λ?= 0.000016 ?? λ?= 1.6 × 10−5?? 4 JAYAM CHEMISTRY ADDA

4. Problem-4 What is the frequency of blackbody radiation if its temperature is 300 Kelvin? Temperature of blackbody radiation = 300 Kelvin Answer Value of Wien constant in Hertz = 0.058 THz / Kelvin According to Wien displacement law, we have; νm= b x T ν?= 0.058 ???/ ? × 300 ? ν?= 17.4 ??? 1 THz (Terahertz) = 1012 Hertz ν?= 1.74 × 1013????? The blackbody radiation falls in the infrared region of the electromagnetic spectrum. 5 JAYAM CHEMISTRY ADDA

5. Problem-5 The temperature of an incandescent electric bulb is 4982 degrees centigrade. Then what is the wavelength of emitted radiation? Answer The temperature of an incandescent electric bulb = 4982 degrees centigrade Converting temperature to Kelvin scale, we have; 49820 C + 273= 5255 K According to Wien displacement law, we have; λmx T = b λ?=b T Value of b = 289 x 10-5mK λ?=289 × 10−5m K 5255 K λ?= 0.05499 × 10−5? = 549.9 ?? 6 JAYAM CHEMISTRY ADDA

6. Problem-7 A heated metal body emits thermal electromagnetic radiation having 4000 nm at 2000 Kelvin. If the body temperature doubles, what will be the wavelength of emitted radiation? The wavelength of the heated metal body = 4000 nm = 4000 x 10-9m Answer Temperature of emitted electromagnetic radiation = 2000 Kelvin According to Wien displacement law, we have; λmx T = b 4000 × 10−9? × 2000 ? = ? ----------------------(1) Upon heating, the temperature of electromagnetic radiation = 4000 Kelvin Again by Wien displacement law, we have; λmx T = b λ?× 4000 ? = ? ---------------------------------- (2) 8 JAYAM CHEMISTRY ADDA

7. On comparing both the equations (1) and (2) we get; (4000 x 10-9m) x 2000 K = λm x 4000 K 4000 × 10−9? × 2000? 4000 ? λ?= λm= 2000 x 10-9m λm= 2000 nm The wavelength of emitted thermal electromagnetic upon increasing temperature is equal to 2000 nm. 9 JAYAM CHEMISTRY ADDA

8. Problem-15 What is the temperature of blackbody radiation if its maximum intense wavelength is at 4500 A0? Provided that k= 1.38 x 10-23j/K. Wavelength of thermal radiation = 4500A0= 4500 x 10-10m Answer According to Wien displacement law, we have; ℎ? λ?? = 4.965? Where, h = Planck constant c= Velocity of light in vacuum (6.626 × 10−34??) (3 × 108? ???−1) 4.965 1.38 × 10−23?/? (4500 × 10−10?) ? = 19.878 × 10−26?? 30832.65 × 10−33??/? ? = 19 JAYAM CHEMISTRY ADDA

9. ? = 0.0006447 × 107? T = 6447 K 20 JAYAM CHEMISTRY ADDA

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