Compute the power output in m W that could be obtained durin

Compute the power output (in m W) that could be obtained during the decay of 1,00 gram of 239_Pu.

The decay reaction is

Pu(239/94) ......> U (235/92) + He (4/2)

The energy is

Q= M(Pu ) - M(U) - M (He)

= (239.05216 u - 235.0439299 u - 4.0026u)(931.5 MeV/c^2/1 u)

= 5.24 MeV

Number of nuclei is

N=(1 g/239 g/mol)*(6.02*10^23)=2.518*10^21

Decay constant is

Lambda=ln(2)/T_1/2 = ln(2) /2.41*10^4 y = 2.876*10^-5 /yr

Decay rate is

R = N*Lambda = 7.24*10^16 decay/yy

Power output is

P=QR= (5.24 MeV)(1.6*10^-13 J/MeV)(7.24*10^16)

= 6.07*10^4 J/yr

= 1.924*10^-3 J/s or

= 1.924 mW