Q16 of 32 Page 462

Suppose, we think of fission of a 5626Fe nucleus into two equal fragments, 2813Al. Is the fission energetically possible? Argue by working out Q of the process. Given m (5626Fe) = 55.93494 u and m (2813Al) = 27.98191 u.

The fission reaction can be given as :

The Q value for this reaction will be given as = [m()-2 × m()] × c2


= (55.93494-2 × 27.98191)u × c2 MeV


= -0.02888u × c2 MeV


= -0.02888 × 931.5


= -26.902MeV


The Q value is negative which suggests this reaction is endothermic, but we know fission reactions are exothermic. Hence, this fission reaction is not energetically possible.


More from this chapter

All 32 →
14

The nucleus 2310Ne decays by β– emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

m (2310Ne) = 22.994466 u


m (2311Ne) = 22.089770 u.

 15

The Q value of a nuclear reaction A + b → C + d is defined by

Q = [ mA + mb– mC – md]c2


Where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.


(i) 11H + 31H → 21H + 21H


(ii) 126C + 126C → 2010Ne + 42He


Atomic masses are given to be


m (21H) = 2.014102 u


m (31H) = 3.016049 u


m (126C) = 12.000000 u


m (2010Ne) = 19.992439 u

 17

The fission properties of 23994Pu are very similar to those of 23592U. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure 23994Pu undergo fission?

 18

A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much 23592U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of 23592U and that this nuclide is consumed only by the fission process.