00:01
In the question it is asking that how much energy is released by 0 .895 kg of deuterium.
00:17
So and other information is given that is mass of neutron is given that is 1 .00866 amu.
00:38
So by using this data i have to find out this energy.
00:45
So first step is write the equation that is that will show the reaction between deuterium.
00:59
So this equation is that is after balancing equation is 1 h2 plus 1 as 2 after taking this reaction this reaction they will will produce that is first helium then neutron zero and one then some energy so some energy is released so this is the reaction now next step is find the energy that is released for these two deuterium atom so what we have to do is first we have to find the mass defect means change in mass so as you see we know the first mass of deuterium that is 1 h2 is equal to 2 .0141 u means and the mass of helium that is m2 is equal to 3 .0164 u and the mass of neutron that is given in the question that is m3 is given 1 .00866 u here that u that u is a unit of mass and you will show that is atomic it is also called that is atomic mass unit.
03:10
Now as i already said i will have to find first mass defect means change in mass so by taking this above reaction i can say that delta m will be equal to 2 times of m1 minus bracket open then m2 plus m3 so put the value of all data so after you play i can say that delta m will be equal to that is 2 times of 2 .01410 minus m2 is that is 3 .01664 plus 1 .m3 is that is 1 .00866.
04:08
Now solve this then i will get the value of change in mass or i can say delta mass defect that is equal to after selling value is that is 0 .0029 u now total energy that is released will be by this two deuterium atom that is equal to 0 .0029 times of 931 .5 here this term is come out from as we know the unit conversion that is 1 amu is equal to 9331m .5 m .ev means mega electron volt.
05:08
So by using this conversion i can say that the total energy that is resist by two deuterium atom is equal to 0 .0 .0029 times of 931 .5.
05:26
So this will give that is equal to 2 .70135 mega electron volt or i can say that 2 .70 mega electron volt.
05:49
So this energy total energy released by two deuterium atom.
05:57
Now next step is first find the number of atom which is given by given amount of the amount of the amount of the atom.
06:12
Of deuterium...