Question

When ²³⁵U fissions, the fission products are not always the same and exactly two neutrons are not always released. An example of one fission reaction is

235U + n 141Cs + 93Rb + 2n

You may want to use the following table of atomic masses:

Table of masses
141Ba	140.9144 u		144Ba	143.9229 u		139Te	138.9347 u
141Cs	140.9196 u		90Kr	89.91952 u		91Kr	90.92344 u
92Kr	91.92615 u		94Zr	93.90632 u		93Rb	92.92157 u
235U	235.0439 u		p	1.00728 u		n	1.00867 u

1. What is the energy released in this fission?

Q =

2. Another example of possible results is

Q =

3. Lets assume that these are the only two fission reactions that take place (they are not), and that they take place at equal rates so the amount of energy released in every fission of ²³⁵U is the average of these two. How many gallons of gas would have to be burned to release the same amount of energy released when one kilogram of ²³⁵U undergoes fission? (A gallon of gasoline releases about 2.0X10⁸J of energy when burned.)

N_gallons =

Answer 1

we know that I amu q to mass produces 93105 mev q energy. . for reaction 20+ mo 'le + 9386+21 = 11235.0439 +1.00867)–(140-9196+99.90157+2X1•00 864) X 9315 men Q-180.76689 MeV 07, Q = |bxto 19 x 106 x180.76689 I .' Q=2.89 XIV J of 180.76619 Mey © for reaumon 2350 +14BA + 92K4+39 Q = (23.5. 0449 +1.00867) - (140:9144 +91.92615+3X1:00861)x93)6 Mey Q=173-3 Mer on, Q = 1.6x1079 x106 X171.3 I : 27728 x1071), w 19-2-7728x10-11.or 173. 3 Mevl on average 1 arom 2350 produce, 23(2-8942.7928 kb. ly. i Payo = 2-8314x10" i l ky q 2350 contains a 60022 x1026 atome 235.0489a i Total Energy produced by I kg 2950, Eq= 2.821x15" (602x56 235.0429 I ET +7.25x16? I , hasoline required to produce thusenergy, v= 7.3% 103 calline TV3•627 xios gallons q gasoline - 2x10 & gallery