According to stefan's law of radiation a black body radiates energy `sigmaT^(4)` from is unit surface area every second where `T` is the surface temperature of the black body and `sigma = 5.67 xx 10^(-8) W//m^(2) K^(4)` is known as Stefan's of as a ball of radius `0.5m` When detonated it reaches temperature of `10^(6)K` and can be treated as a black body Estimate the power it radiates .

Explanation

(i) Radius of the ball,`r=0.5m" "sigma=5.67xx10^(-8)Wm^(-2)K^(-4),T=10^(6)K`
Surface area of nuclear weapon, `A=4pir^(2)=4xx3.14xx(0.5)^(2)=3.14m^(2)`
`therefore` Power radiated , `p=(sigmaT^(4))A=5.67xx10^(-8)xx(10^(6))^(4)xx3.14=1.8xx10^(17)W`
(ii) Energy radiated , `E=Pxxt`
`=1.8xx10^(17)Wxx1s=1.8xx10^(17)J`
`Q=10% "of"E=1.8xx10^(16)J`
Also , `Q=mC_(w)DeltaT+mL_(V)=m(C_(W)DeltaT+L_(V))`
Here `C_(W)=4186Jkg^(-1)K^(-1)`
`L_(V)=22.6xx10^(5)Jkg^(-1)`
`DeltaT=100-30=70^(@)C=70K`
`thereforem=(Q)/((C_(W)DeltaT+L_(V)))=(1.8xx10^(16))/(4186xx70+22.6xx10^(5))=(1.8xx10^(16))/(25.53xx10^(5))=7xx10^(9)kg`
(iii) `becausep=(U)/(C)`
The radiation spread in an area of `4pir^(2)`
Here, `r=1km=10^(3)` m
Momentum imparted per unit time on a unit area at a distance r,
`=((U//C))/(4pir^(2)t)=(U)/(4piCr^(2)t)=4piCr^(2)=(1.8xx10^(17))/(4xx3.14xx3xx10^(8)xx10^(6))=47.7Nm^(-2)`