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Nuclear Fission and Nuclear Fusion are the core components of Modern Physics. Questions are often picked up from these topics in the JEE. It is therefore important for the JEE aspirants to master these topics to remain competitive in the exam. The major benefit of mastering this topic is that the numerical framed on it are easy as well as scoring. Basic concepts like calculation of energy released in fission or numerical on nuclear fusion are some of the pet questions asked. We discuss here some examples like calculating the released energy in a nuclear reaction……….
Q1. In the following nuclear reaction _{6}C^{11} —> _{5}B^{11} + β + X. What does X stand for?
(A) A neutron (B) A neutrino
(C) An electron (D) A proton
Solution:
The given reaction is _{6}C^{11} —> _{5}B^{11} + β + X We know that in beta decay, a neutrino is accompanied with the emission of a positron. Hence here the correct answer is the neutrino.
Q2. If 200 MeV energy is released in the fission of a single U^{235} nucleus, the number of fissions required per second to produce 1 kilowatt power shall be (given 1 eV = 1.6 × 10^{–19} J)
(A) 3.125×10^{13} (B) 3.125×10^{14}
(C) 3.125×10^{15} (D) 3.125×10^{16}
The fission of a single U^{235} nucleus releases 200 MeV of energy. Energy released in the fission is given by the formula E = 200 MeV = 200 × 10^{6 }× 1.6 × 10^{–19 }= 3.2 × 10^{–11} Joule P = nE/t => n/t = P/E = 10^{3} / 3.2 × 10^{–11} = 3.125×10^{13} Hence, the correct option is (A).
Q3. M_{P} = 1.008 a.m.u., M_{N} = 1.009 a.m.u. and _{2}He^{4} = 4.003 a.m.u. then the binding energy ofα-particle is
(A) 21.4 MeV (B) 8.2 MeV
(C) 34 MeV (D) 28.8 MeV
We are given the values of M_{P}, M_{N} and _{2}He^{4}. We know the formula for finding the binding energy is given by BE = 2M_{p} + 2M_{n} – M_{a} Hence, substituting the values in the given formula BE = 2 × 1.008 + 2 × 1.009 – 4.003 = 0.031 amu = 0.031 × 931 = 28.8 MeV So, the correct answer is (D).
Q4. Energy released in fusion of 1 kg of deuterium nuclei
(A) 9×10^{13 }J (B) 6×10^{27 }J
(C) 2×10^{7 }KwH (D) 8×10^{23 }MeV
First we consider the fusion reaction of deuterium. The reaction is _{1}H^{2} + _{1}H^{2} —> _{2}He^{3 }+ _{0}n^{1}+3.27 MeV So applying the formula for energy released and then substituting the values we get, E = 6.02 × 10^{23} × 10^{3} × 3.27 × 1.6 × 10^{–13} / 2 × 2 => E ∼ 9 × 10^{13} J, so this gives (D) as the correct option.
Q5. The energy produced in the sun is due to
(A) Fission reaction (B) Fusion reaction
(C) Chemical reaction (D) Motion of electrons and ions
It is a known fact that the source of the huge solar energy is the fusion of lighter nuclei. Fusion of hydrogen nuclei into helium nuclei is continuously taking place in the plasma, with the continuous liberation of energy. Therefore energy produced in the sun is due to fusion reaction. Obviously, (B) is the correct option.
Q6. Atomic number of a nucleus is Z, while its mass number is M. What will be the number of neutrons in its nucleus?
(A) M (B) Z
(C) (M–Z) (D) (M+Z)
Clearly, the question demands the relation between the atomic number and the mass number. We know the formula M= Z + N, where M is the mass number, Z is the atomic number and N is the number of neutrons. Hence, N= M-Z. the correct option is (C). Watch this Video for more reference
Q7. In uncontrolled chain reaction, the quantity of energy released, is
(A) Very high (B) very low
(C) Normal (D) first (A) to (B)
In an uncontrolled chain reaction, a particular fission may produce more than one neutron and these neutrons cause further fissions which increases the number of fissions very rapidly. This is indeed an extremely fast process and the whole substance is fissioned within a few moments. This releases a huge amount of energy and so the correct option is (A).
Q8. The net force between two nucleons 1 fm apart is F_{1} if both are protons, F_{2} if both are neutrons, and F_{3} if one is a neutron and the other is a proton.
(A) F_{1} < F_{2} < F_{3} (B) F_{2} < F_{1} < F_{3}
(C) F_{1} < F_{2} = F_{3} (D) F_{1} = F_{2} < F_{3}
The nuclear force of interaction between any pair of nucleons is identical i.e. force between two neutrons (F_{2}) equals the force between neutron and proton (F_{3}). However, between two protons net force is equal to the resultant of nuclear force between them (attractive in nature) and electrostatic force between them (repulsive in nature). Hence F_{1} is a value lesser than F_{2} and F_{3}. So F_{1} < F_{2} = F_{3}. Hence, the correct option is (C).
The examples listed above are just to give you an idea of the types of questions framed on the topic. askIITians offers solved numerical on nuclear fission as well as fusion in its free study material. Concepts like how to calculate energy released from the solar fusion of deuterium or energy released from the fusion of deuterium and proton have also been covered in the material.
Click here for the Detailed Syllabus of IIT JEE Physics.
Get an idea about the kinds of questions asked by referring the Model Papers of Previous Years.
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