According to Wein's law:
1. ${\mathrm{\lambda }}_{\mathrm{m}}\mathrm{T}$= constant                 

2. $\frac{{\mathrm{\lambda }}_{\mathrm{m}}}{\mathrm{T}}$= constant

3. $\frac{T}{{\mathrm{\lambda }}_{\mathrm{m}}}$= constant                 

4. $\mathrm{T}+{\mathrm{\lambda }}_{\mathrm{m}}$= constant

A black body has a maximum wavelength at a temperature of 2000 K. Its corresponding wavelength at temperatures of 3000 K will be: 

A black body at \(200\) K is found to emit maximum energy at a wavelength of \(14\) \(\mu \)m. When its temperature is raised to \(1000\) K, the wavelength at which maximum energy is emitted will be:

A piece of iron is heated in a flame. If it becomes dull red first, then becomes reddish yellow, and finally turns to white hot, the correct explanation for the above observation is possible by using:

If ${\lambda }_m$ is the wavelength, corresponding to which the radiant intensity of a block is at its maximum and its absolute temperature is T, then which of the following graph correctly represents the variation of T?

1.		2.
3.		4.

Three stars \(A,\) \(B,\) and \(C\) have surface temperatures \(T_A,~T_B\)and \(T_C\) respectively. Star \(A\) appears bluish, star \(B\) appears reddish and star \(C\) yellowish. Hence,

The plots of intensity versus wavelength for three black bodies at temperatures ${\mathrm{T}}_1$, ${\mathrm{T}}_2$ and ${\mathrm{T}}_3$ respectively are as shown. Their temperatures are such that:
         

The energy distribution E with the wavelength $\left(\mathrm{\lambda }\right)$ for the black body radiation at temperature T Kelvin is shown in the figure. As the temperature is increased the maxima will:
     

A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at a wavelength of 250 nm is U₁, at a wavelength of 500 nm is U₂ and that at 1000 nm is U₃. Given Wien's constant $b=2.88\times {10}^6$ $nm-\mathrm{K,\; which }$of the following is correct?
1. U₃=0       
2. U₁>U₂
3. U₂>U₁     
4. U₁=0
 

If the temperature of the sun becomes twice its present temperature, then: