Question

The temperature of an ordinary electric bulb is around $300K$. At what wavelength will it radiate maximum energy? Given $b = 0.288cmK$ .
A) $8600\mathop {\rm A}\limits^{\text{o}} $
B) $9600\mathop {\rm A}\limits^{\text{o}} $
C) $4655\mathop {\rm A}\limits^{\text{o}} $
D) $1000\mathop {\rm A}\limits^{\text{o}} $

Answer 1

Hint: According to Wein’s displacement law the wavelength corresponding to maximum emission of radiation is inversely proportional to the temperature of the body. In the graph formed between Temperature and the wavelength of radiation emitted by the body the maximum wavelength tends to increase with decrease in the temperature. Here we convert a unit of Wein’s constant from $cmK$ to $mK$ .

Formula used:
${\lambda _{\max }} = \dfrac{b}{T}$
Where,
${\lambda _{\max }}$ is the wavelength peak,
$b$ is the constant of proportionality and
$T$ is the absolute temperature.

Complete step by step solution:
Wien's displacement law states that for a black-body radiation the curve of different temperatures will reach its peak at different wavelengths that are inversely proportional to the temperature.
According to the formula given above the temperature of the electric bulb happens to be 3000 kelvins from the question and on converting Wein’s constant i.e., b from $cmK$ to $mK$ we have $b = 0.288cmK$ .
So, by Wien's Displacement law,
$
  {\lambda _{\max }} = \dfrac{b}{T} \\
   \Rightarrow {\lambda _{\max }} = \dfrac{{0.00288}}{{3000}} \\
   \Rightarrow {\lambda _{\max }} = 9.6 \times {10^{^{ - 7}}}m \\
   \Rightarrow {\lambda _{\max }} = 9.6 \times {10^{ - 7}} \times {10^{10}}\mathop {\rm A}\limits^{\text{o}} \\
   \Rightarrow {\lambda _{\max }} = 9600\mathop {\rm A}\limits^{\text{o}} \\
 $
On solving we get the value of lambda is $9600\mathop {\rm A}\limits^{\text{o}} $ which is the value of wavelength of maximum energy radiation emitted from the bulb of temperature of $3000K$ .

Note:
The wavelength so came out to be $9600\mathop {\rm A}\limits^{\text{o}} $ does not lie in the visible range therefore the radiation emitted by the bulb cannot be seen by our eyes. Since we know for any radiation to be visible it must lie in the range of $4000{\text{ to }}7000\mathop {\rm A}\limits^{\text{o}} $ . This shows that the bulb radiates infrared radiation which ranges above $7000\mathop {\rm A}\limits^{\text{o}} $.