According to Stefan's law of radiation,a black body radiates energy $\sigma T^4$ from its unit surface area every second,where $T$ is the surface temperature of the black body and $\sigma = 5.67 \times 10^{-8} \, W m^{-2} K^{-4}$ is known as Stefan's constant. $A$ nuclear weapon may be thought of as a ball of radius $0.5 \, m$. When detonated,it reaches a temperature of $10^6 \, K$ and can be treated as a black body.
$(a)$ Estimate the power it radiates.
$(b)$ If the surroundings have water at $30 \, ^\circ C$,how much water can $10 \%$ of the energy produced evaporate in $1 \, s$? $[S_W = 4186 \, J kg^{-1} K^{-1}$ and $L_v = 22.6 \times 10^5 \, J kg^{-1}]$
$(c)$ If all this energy $U$ is in the form of radiation,the corresponding momentum is $p = U/c$. How much momentum per unit time does it impart on a unit area at a distance of $1 \, km$?

(N/A) Power emitted according to Stefan-Boltzmann law is $P = \sigma A T^4$.
Given $A = 4 \pi R^2 = 4 \times 3.14 \times (0.5)^2 = 3.14 \, m^2$.
$P = 5.67 \times 10^{-8} \times 3.14 \times (10^6)^4 = 1.78 \times 10^{17} \, W \approx 1.8 \times 10^{17} \, J/s$.
$(b)$ Energy available for evaporation $Q = 10 \% \text{ of } P = 0.1 \times 1.78 \times 10^{17} = 1.78 \times 10^{16} \, J$.
Heat required to evaporate mass $m$ of water: $Q = m S_W \Delta T + m L_v$.
$1.78 \times 10^{16} = m [4186 \times (100 - 30) + 22.6 \times 10^5]$.
$1.78 \times 10^{16} = m [2.93 \times 10^5 + 22.6 \times 10^5] = m [25.53 \times 10^5]$.
$m = \frac{1.78 \times 10^{16}}{25.53 \times 10^5} \approx 6.97 \times 10^9 \, kg$.
$(c)$ Momentum per unit time per unit area is radiation pressure $Pr = \frac{P}{4 \pi r^2 c}$.
$Pr = \frac{1.78 \times 10^{17}}{4 \times 3.14 \times (1000)^2 \times 3 \times 10^8} = \frac{1.78 \times 10^{17}}{3.77 \times 10^{15}} \approx 47.2 \, N/m^2$.