The elecron curium `_(96)^(248)` On has a mean life of `10^(13)` second ist pirmary dacay mode with a probilly of `8%` and the letter with a probillty of `92%` Each fission released `200 MeV` of energy . The masses involved in a - dacay are as follows `_(96)^(248) Cm = 248.072220 u, _(94_^(244) Pu = 244.064100 u and _(2)^(4) He = 4 .002603 u ` calculate the power output from a sample of `10^(20) Cm atom (1 u = 931 MeV//e^(2))`

Answer

A, B, C

Explanation

The reaction involed in a - decay is
`_(96)^(248) Cm rarr _(94)^(244) Pu + _(2_^(4) He`
Mass defect
` delta m = mass of _(96)^(248) Cm - mass of _(94)^(244) Pu - mass of _(2_^(4) He`
`= (248.072220 - 244.064100 - 4.002603) u `
`= 0.005517 u`
Therefore , energy released in a - decay will be
`E_(a) = (0.005517 xx 931) MeV = 5.136 MeV` ltvbrgt similarly `E_(fassion) = 200 MeV `(given)
mass of is given as `t_(mean) = 10^(13) s = (1)/(lambda)`
`:.` Disingration constant `lambda = 10^(-13) s^(-1)`
rate of decay at the moment when number of nuclei are `10^(20)` is
`(dN)/(dt) = lambda N = (10^(-13))(10^(20)) = 10^(7) dps`
Therefore energy released , `8%` are in fission and `92%` are in a - decay
`= (0.08 xx 10^(7) xx 200 + 0.92 xx 10^(7) xx 5.1 36)MeV`
= 2.074 xx 10^(8) MeV`
:. power of output (in watt) = Energy released per second (J//s)
`= (2.074 xx 10^(8))(1.6 xx 10^(-13))`
power output `= 3.32 xx 10^(-5) watt`.