1. (JEE-Advanced
2015) A fission reaction is given by
92U236
→ 54Xe140 + 38Sr94 +X +Y, where X
and Y are two particles. Considering 92U236
to be at rest, the kinetic energies of the products are denoted by Kxe,
KSr, KX (2 MeV) and KY (2 MeV) respectively.
Let the binding energies per nucleon of 92U236, 54Xe140
and 38Sr94 be 7.5 MeV, 8.5 MeV and 8.5 MeV respectively.
Considering different conservation laws, the correct option(s) is (are)
a) X=n, Y=n KSr = 129 MeV Kxe = 86
MeV
b) X=p, Y=e- KSr = 129 MeV Kxe
= 86 MeV
c) X=p, Y=n KSr = 129 MeV Kxe = 86
MeV
d) X=n, Y=n KSr = 86 MeV Kxe = 129
MeV
Answer : a) X=n, Y=n KSr = 129 MeV Kxe = 86 MeV
Solution:
92U236 → 54Xe140 + 38Sr94
+0X1 +0Y1
X = Y = n (same)
Q = (236×7.5)-{(140×8.5)+(94×8.5)} = 219 MeV
In options A and D, Energy and charge conservation is
followed.
Q = Kxe + KSr + KX + KY
= 129 + 86+ 2 + 2 = 219 MeV
Two particles will have same mass. But lighter one
will have higher KE.
In option D, PXe > PSr + PX
+ PY. So conservation of momentum will not hold.
2. A reactor is developing nuclear energy at the rate
of 32,000 kilowatts. How many atoms of U-235 undergo fission per second? How
many Kg of U-235 would be used up in 1000 hours of operation? Assume an average
of 200 MeV released per fission. Take Avogadro number as 6 × 1023 /g
mole and 1 MeV = 1.6 × 10-13 joule.
Solution:
Energy released per fission = 200 MeV = 200 × 1.6 × 10-13
J
= 320 × 10-13 joule
Let N be the number of fissions per second.
Total Energy released = Number of fissions per second ×
Energy per fission
32000 × 103 J/s = 320 × 10-13 ×
N
N = 1018 fissions per second
In one second 1018 fissions.
For 1000 hours, N1 = 1018 × 1000
× 3600 = 36 × 1023
One gram atom =
6 × 1023 atoms.
M = (36 × 1023 × 235)/(6 × 1023)
= 1410 gm = 1.41 kg
3. Calculate number of fissions required per second to
produce 1 kilowatt power in reactor. Assume an average of 200 MeV released per
fission. Take 1 MeV = 1.6 × 10-13 joule.
Solution:
Power = 1 KW = 103 J/s
Energy released per fission = 200 MeV = 200 × 1.6 × 10-13
J
= 320 × 10-13 joule
Power = nE/t
Number of fission required per second : (n/t) = P/E = (103 / 320 × 10-13)
= 3.125 × 1013
