A black body at a temperature of 2600 K has the wavelength corresponding to maximum emission 1200 mathringA. Assuming the moon to be perfectly black body the temperature of the moon, if the wavelength corresponding to maximum emission is 5000 mathringA is

Question

A black body at a temperature of 2600 K has the wavelength corresponding to maximum emission 1200 mathringA. Assuming the moon to be perfectly black body the temperature of the moon, if the wavelength corresponding to maximum emission is 5000 mathringA is (A) 7800 K (B) 6240 K (C) 5240 K (D) 3640 K

Tardigrade Admin · Accepted Answer

From Wien law λ1 T1 =λ2 T2 T2 =(λ1 T1/λ2) =(1200 × 2600/5000) T2 =6240 K

Q. A black body at a temperature of $2600 K$ has the wavelength corresponding to maximum emission $1200 \overset{˚}{A}$ . Assuming the moon to be perfectly black body the temperature of the moon, if the wavelength corresponding to maximum emission is $5000 \overset{˚}{A}$ is

7800 K

6240 K

5240 K

3640 K

From Wien law
$λ_{1} T_{1} = λ_{2} T_{2}$
$T_{2} = \frac{λ _{1} T _{1}}{λ _{2}}$
$= \frac{1200 \times 2600}{5000}$
$T_{2} = 6240 K$

Q. A black body at a temperature of 2600K has the wavelength corresponding to maximum emission 1200A˚. Assuming the moon to be perfectly black body the temperature of the moon, if the wavelength corresponding to maximum emission is 5000A˚ is