Calculate Power of a Nuclear Reactor | IIT JEE 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:

Q1. The mass number of a nucleus is

(A) Always less than atomic number

(B) Always more than 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).

Q2. In which of the following decays, the element does not change?

(A) β – decay (B) α – 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).

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Q3. 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.
(M_N = 14.00307 amu, M_He = 4.00260 amu and M_O
= 16.99914 amu, M_H = 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).

Q4. In the carbon cycle of fusion
(A) Four ₁H¹ fuse to form ₂He⁴and two positrons
(B) Four₁H¹ fuse to form ₂He⁴and two electrons
(C) Two ₁H² fuse to form ₂He⁴
(D) Two ₁H² fuse to form₂He⁴ 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 ₂He⁴and two positrons. So, (A) is the correct option.

Q5. 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 U²³⁵is 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.

Q6. 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.

Q7. 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.