Nuclear Fission Problems

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 K_xe, K_Sr, K_X (2 MeV) and K_Y (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 K_Sr = 129 MeV K_xe = 86 MeV

b) X=p, Y=e^- K_Sr = 129 MeV K_xe = 86 MeV

c) X=p, Y=n K_Sr = 129 MeV K_xe = 86 MeV

d) X=n, Y=n K_Sr = 86 MeV K_xe = 129 MeV

Answer : a) X=n, Y=n K_Sr = 129 MeV K_xe = 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 = K_xe + K_Sr + K_X + K_Y = 129 + 86+ 2 + 2 = 219 MeV

Two particles will have same mass. But lighter one will have higher KE.

In option D, P_Xe > P_Sr + P_X + P_Y. 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