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A radioactive with half life T=693.1 day...

A radioactive with half life `T=693.1` days. Emits `beta`-particles of average kinetic energy `E=8.4xx10^(-14)` joule. This radionuclide is used as source in a machine which generates electrical energy with efficiency `eta=12.6%`. Number of moles of the nuclide required to generate electrical energy at an initial rate is `P=441 KW`, is `nxx10^(n)` then find out value of `(n)/(m) (log_(e )2=0.6931)N_(A)=6.023xx10^(23)`

Text Solution

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The correct Answer is:
`2`
Let number of moles of the radio nuclide be `'n'`
Number of nclide `=n.N`
since, half of the nuclide is `T`, therefore, its decay constant `lambda=(log2)/(T)`
`:.` Activity of rate of decay, `A=lambda("mole").(N)=(("mole").Nlog2)/(T)`
Energy released per decay `=` average kinetic energy `E` of `beta`-particles
`:.` Rate of release of the energy `=AE`
But efficiency of conversion of this energy into electrical energy is `eta`. therefore, rate of generation of electrical energy is
`P= eta A E=(etaEnNlog2)/(T)`
or `"mole"=(TP)/(etaEnNlog_(e )2)=6000`
`=6xx10^(3)rArrm=3(n)/(m)=2`
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