1 Ci = 3.7*10^10 disintegration per second. For each disintegration one alpha particle with 1 MeV or [1.6*10^-13] J. So the required power in watts (J/s) = [3.7*1.6]*10^-3 W = 5.92 mW
What is the power (in watts) of a 1 gram sample of 238 94 Pu ? Assumi8ng that each nucleus emits only one alpha particle of the same energy answer in watts = [6.023*10^23](1/238)(1.6*10^-13) = 4*10^8 = 400 MW
How much 241 95 Am would it take to power a 60 W light bulb? Assume the heat from the decay is converted to electricity with an efficiency of 15%.
Let m gram of Am241 be required assuming similar things about decay of this nucleus as well we have 60 = 0.15*m*[(6.023*1.6)/241]*10^10 or m = [(60*241)/[6.023*1.60.15)]*10^-10 gram or m = 1microgramme I suggest you Unit Converter with help of this you get more accurate results your problems.
1 Ci = 3.7*10^10 disintegration per second. For each disintegration one alpha particle with 1 MeV or [1.6*10^-13] J. So the required power in watts (J/s) = [3.7*1.6]*10^-3 W = 5.92 mW
ReplyDeleteWhat is the power (in watts) of a 1 gram sample of 238
94 Pu ? Assumi8ng that each nucleus emits only one alpha particle of the same energy answer in watts = [6.023*10^23](1/238)(1.6*10^-13) = 4*10^8 = 400 MW
How much 241
95 Am would it take to power a 60 W light bulb?
Assume the heat from the decay is converted to electricity with
an efficiency of 15%.
Let m gram of Am241 be required assuming similar things about decay of this nucleus as well we have
60 = 0.15*m*[(6.023*1.6)/241]*10^10 or m = [(60*241)/[6.023*1.60.15)]*10^-10 gram or
m = 1microgramme I suggest you Unit Converter with help of this you get more accurate results your problems.