Solved Examples on Power of a Nuclear Reactor
Nuclear Reactor is an important topic of IIT JEE syllabus. The topic is quite simple and easily fetches 3-4 questions in the JEE. With a bit of hard work it becomes easy to master this topic and in fact, it is vital to have a good hold on it to remain competitive in the IIT JEE. The questions asked in JEE may range from the simple ones like estimating number of fissions per second to difficult ones like the nuclear power calculation. Some sample problems with solution of nuclear energy fission and fusion and on nuclear reactor have been discussed here:
1. The mass number of a nucleus is
(A) Always less than atomic number
(B) Always more than atomic number
(C) Equal to atomic number
(D) Some times more than and sometimes equal to atomic number.
Solution: We know that the mass number of a nucleus represents the number of nucleons (neutrons + protons), and the atomic number represents the number of neutrons in the nucleus. So, when there are no neutrons in the nucleus, only then the atomic number equals the mass number. Hence, the correct option is (D).
2. In which of the following decays, the element does not change?
(A) β – decay (B) α – decay
(C) Positive– decay (D) γ – decay
Solution: We know that γ ray has no charge and no mass. Hence, it is during the emission of γ ray that there is no change in atomic number or mass number. So, the correct option is (D).
Watch this video to view more on alpha and beta rays
3. In the reaction 7N14 + 2He4 —> 8O17 + 1H1 the minimum energy of the α-particle is
(A) 1.21 MeV (B) 1.62 MeV
(C) 1.89 MeV (D) 1.96 MeV.
(MN = 14.00307 amu, MHe = 4.00260 amu and MO
= 16.99914 amu, MH = 1.00783 amu and 1 amu = 931 MeV)
Solution: The given reaction is 7N14 + 2He4 —> 8O17 + 1H1
We first calculate the total mass of reactants as well as products
Total mass of reactants = 18.00567 amu
Total mass of products = 18.00697 amu
Mass defect = 18.00697 – 18.00567 = 0.0013 amu
Energy (E) = 931 (0.0013) = 1.2103 MeV
Hence the correct option is (A).
4. In the carbon cycle of fusion
(A) Four 1H1 fuse to form 2He4 and two positrons
(B) Four 1H1 fuse to form 2He4 and two electrons
(C) Two 1H2 fuse to form 2He4
(D) Two 1H2 fuse to form 2He4 and two neutrons
Solution: The carbon cycle of fusion is given by this equation
41H1 —> 2He4 + 2. +1e0 + Energy
Hence, it is clear that in this equation four 1H1 fuse to form 2He4 and two positrons. So, (A) is the correct option.
5. In each fission of U235, 200 MeV of energy is released. If a reactor produces 100 MW power, then the rate of fission in it will be
(A) 3.125 × 1018 per minute
(B) 3.125 × 1017 per second
(C) 3.125 × 1017 per minute
(D) 3.125 × 1018 per second
Solution: The energy released in every fission of U235 is given in the question. We know the formula P = nE/t.
Hence, substituting the values in this formula, we get
n/t = P/E = 100×106 / 200[1.6×1013]
= 3.125 × 1018 /sec
SO (D) is the correct option.
6. To generate a power of 3.2 MW, the number of fissions of U235 per minute is (Energy released per fission = 200 MeV, 1 eV = 1.6 ´ 10–19J)
(A) 6×1018¡ (B) 6×1017
(C) 1017 (D) 6×1016
Solution: The power of reactor P = nE/t
Here, ‘n’ denotes the number of fissions,‘t’ denotes the time and ‘E’ is the energy released per fission.
∴ 3.2 × 106 = n(200×106)(1.6×10–19) / 60
=> n = 6 × 1018. This gives (A) as the correct option.
7. If in nuclear reactor using U235 as fuel, the power output is 4.8 MW, the number of fissions per second is (Energy released per fission of U235 = 200 MeV watts, e eV = 1.6 ´ 10–19 J)
(A) 1.5×1017 (B) 3×1019
(C) 1.5×025 (D) 3×1025
Solution: The power output is given to be 4.8 MW. This may be represented as
P = 4.8 MW = 4.8×106 W
The power of a nuclear reactor is given by the formula P = nE/t
∴ n/t = P/E = 4.8×106 / (200)(1.6×10–13) = 1.5 × 1017
So this gives (A) as the correct option.
It is vital for the JEE aspirants to master this topic and practice nuclear fission problems solutions. askIITians offers solved problems about nuclear reaction and nuclear reaction examples.