A black body is at a temperature of `5760 K`. The energy of radiation emitted by the body at wavelength `250 nm` is `U_(1)` at wavelength `500 nm` is `U_(2)` and that at `1000 nm` is `U_(3)`. Wien's consant, `b = 2.88 xx 10^(6) nmK`. Which of the following is correct?

Options

A. `U_1 gt U_2`
B. `U_2 gt U_1`
C. `U_1 = 0`
D. `U_3 = 0`

Answer

B

Explanation

Maximum amount of emitted radiation correspond to `lamda_m = b/T`
`lamda_m = (2.88 xx 10^6 nmK)/(5760 K) = 500 nm `

`250 nm to U_1`
`500 nm to U_2`
` 1000 nm to U_3`
` therefore U_3 lt U_2 gt U_1`