00:02
Hi everyone, in the first part it is given, calculate the energy released in kilowatt hour if 1 kg of plutonium 239 undergoes complete fission and energy per fission is 200 mega -lequence bolt.
00:41
Let us solve it.
00:43
Number of plutonium nuclei first we will calculate.
00:46
Mass of plutonium atomic weight into avogazard number.
00:51
Mass is given 1 kg.
00:53
Plutonium is 0 .239 kg into avogadra number 6 .023.
01:03
So it is to be 3 .324.
01:19
10 to the power 25.
01:23
So total energy release.
01:26
Is called to number of nuclei into energy per fission that is 3 .324 10 to the power 25 into 200 mega electron bolt so it is to be 5 .04 into 10 to the power 26 mega electron bolt let me correct here it is 10 to the power 24 here also we have to convert in kilowatt hour so it is to be 5 .04 10 to the power 26 mega electron bolt into 4 .44 10 to the power minus 20 kilowatt hour per mega electron bolt.
02:40
So you will get 2 .24 into 10 to the power 7 kilo watt hour.
02:49
Now in b part we have to calculate energy release in deuterium fusion reaction q in 1 h2 plus 1h3 is equal to 2h4 plus neutron.
03:13
So energy release will be equal to mass effect of the reaction into 931 .56 mega electron bolt.
03:23
Mass effect is mass of 1h2, mass of 1h3, mass of 2, mass of 2h3, mass of 2h3, mass of 2h3 ,000...