Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are same. The two bodies emit total radiant power at the same rate. The wavelength `lambda_B` corresponding to maximum spectral radiancy from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by 1.0 `mum`. If the temperature of A is 5802 K, calculate (a) the temperature of B, (b) wavelength `lambda_B`.

Options

A. the temperature of B is 1934 K
B. `lamda_b = 1.5 mu m`
C. the temperature of B is 11604 K
D. the temperature of B is 2901 K

Answer

A, B

Explanation

To Calculate temperature of B
power of body `A= e_(A) sigma T_(A) ^(4) xx ` ( Area )
Power of body `,B =e_(B ) sigma T_(B )^(4) xx` ( area )
the two power are equal
` therefore e_B sigma T_(B )^4 xx `(area ) `=e_(A) sigma T_(A ) ^(4) xx ` (area)
` or T_(B) ^(4) = ((e_A)/(e_(B ))T_(A)^(4) or T_(B)^(4) = ((0.01 )/( 0.81 )) xx (5802 )^4`
` or T_(B )^(4) = ((1)/(3))^4 xx ( 5802)^4 or T_B = ( 5802)/( 3)`
`or T_B = 1934 K`
` therefore ` Option (a) is correct
`(d) ` According to wien.s displacement law
` lamda T=` constant ` therefore lamda_A T_A = lamda_B T_(B)`
` or (lamda_A)/( lamda_B) =(T_B )/(T_A) or (lamda_A)/(lamda_B) = ( 1934 )/(5802 )`
`or (lamda_A)/(lamda_B) -1/3 or lamda_B = 3 lamda_A`
given `lamda_B - lamda_A = 1.0 xx 10^(-6) m`
or ` 3 lamda_A - lamda_A = 10^(-6) or 2 lamda_A = 10^(-6)`
`or lamda_A = 0.5 xx 10^(-6) and lamda_B = 3 xx 0.5 xx 10^(-6)`
` or lamda_B = 1.5 xx 10^(-6) m`
hence option (B) is j correct
options (c ) and (d) are incorrect as option (a) is correct hence option (a) and (b) are correct