## Solved Examples on Nuclear Fission Fusion Binding Energy

We have already discussed the topics of Nuclear Fission and Fusion. These heads lay the groundwork for various topics of radioactivity and assume great importance. The JEE often asks numerical questions from these topics. We discuss here some of the questions on Nuclear Fission and Fusion on the basis of Binding energy.

1. 1 atomic mass unit is equal to(A) 1/12 (mass of one C-atom)

(B) 1/16 (mass of O

_{2}-molecule)(C) 1/14 (mass of N

_{2}-moleucle)(D) 1/25 (mass of F

_{2}-moleucle)

Solution:We know that the atomic mass unit is defined as 1/12th of the mass of one_{6}C^{12}atom. Hence, here, (A) is the right answer.

2. For the construction of nuclear-bomb which of the following substances is taken?(A) U-235 (B) Pu-239

(C) F-14 (D) both (A) and (B)

Solution: For the construction of a nuclear bomb, we need substances which are easily fissionable. Hence, we know that uranium(U^{235})and plutonium (Pu)are easily fissioned by neutrons and hence (D) is the correct option.

3. Energy obtained when 1 mg mass is completely converted to energy is(A) 3×10

^{8 }J (B) 3×10^{10 }J(C) 9×10

^{13 }J (D) 9×10^{15 }J

Solution: The question requires the use of the formula of energy.

We know the energy is given by E =mc^{2}= (1 × 10

^{–3}) (3×10^{8})^{2}

^{ }= 9 × 10^{13}This gives (C) as the correct option.

4. 1 MeV energy is(A) 1.6×10

^{–19 }J (B) 1.6×10^{–16 }J(C) 1.6×10

^{–13 }J (D) 1.6×10^{–11 }J

Solution: we know that the formula for energy is1eV = 1.6 × 10

^{19}J= 1.6 × 10

^{–13}J. Hence, this is the same as (C).

5. Nuclear fission and fusion can be explained on the basis of(A) Conversion of energy principle

(B) Einstein mass-energy equivalence relation

(C) Binding energy per nucleon variation with nucleon number

(D) Variation of mass with increasing atomic nucleus

Solution:We have already discussed that both Nuclear Fission and Fusion are processes in which mass is converted into energy. Hence, out of the given four options (B) seems to be the most appropriate one. Hence, nuclear fission and fusion can be explained on the basis of Einstein mass-energy equivalence relation.

6. If 1 gm hydrogen is converted into 0.993 gm of helium, in a thermonuclear reaction, the energy released in the reaction is(A) 63×10

^{7 }J (B) 63×10^{10 }J(C) 63×10

^{14 }J (D) 63×10^{20 }J

Solution: We know the formula for calculation of energy released in the reaction.

Energy E =Δmc^{2}= (m_{1}– m_{2}) c^{2}= {(1×10

^{–3}) – (0.993×10^{–3})} × (3 × 10^{8})^{2}= (7 × 10

^{–6}) × (9 × 10^{16}) = 63 × 10^{10}J.Hence, (B) is the correct option.

7. In the given particles, which of the following is stable?(A) electron (B) proton

(C) positron (D) neutron

Solution:Since all the particles except neutron have some charge while neutron does not carry any charge, hence neutrons are the most stable. This gives (D) as the correct option.View the following video

8. 1 a.m.u. is equal to(A) 1 gm (B) 4.8×10

^{–10 }esu(C) 6.023×10

^{23 }gm (D) 1.66×10^{–27 }kg

Solution:One atomic mass unit (a.m.u.) is defined as the mass of 1/12^{th}of the isotope of carbon atom (_{6}C^{12}) and is numerically equal to = 1.66 × 10^{–27}kg. so the answer is (D).

9. Hydrogen bomb is based on the phenomenon of(A) nuclear fission (B) nuclear fusion

(C) radioactive decay (D) none of these

Solution:let us discuss the concept of hydrogen bomb in brief. In a hydrogen bomb, first a uranium fission bomb is exploded which produces high temperature of pressure. As a result, the heavy hydrogen nuclei come extremely close to each other and finally fuse, releasing huge amount of energy. Thus, nuclear fission is the reason behind the liberation of this huge amount of energy. The answer is option (B).

10. When two deuterium nuclei fuse together in addition to tritium, we

get a(A) neutron (B) proton

(C) α-particle (D) deuteron

Solution:The fusion reaction of two deuterium nuclei is

_{1}H^{2}+_{1}H^{2}——>_{1}H^{3}+_{1}H_{1}+ QHence, in this reaction besides a tritium we get a proton. So the answer is (B).

11. In nuclear reaction there is conservation of(A) energy only (B) mass, energy and momentum

(C) mass only (D) momentum only

Solution:the nuclear reaction involves the conservation of all three mass, energy as well as momentum. So, the correct option is (B).

12. An atomic power reactor furnace can deliver 300 MW. The energy released due to fission of each of uranium atom U^{238}is 170 MeV. The number of uranium atoms fissioned per hour will be(A) 5×10

^{15}(B) 10×10^{20}(C) 40×10

^{21}(D) 30×10^{25}

Solution:We use the formulae for calculation of fissions per second and per hour.The number of atoms fissioned per second is

= 3 × 10

^{8}/ 27.2 × 10^{–12}= 3× 10^{20}/ 27.2No. of Atoms fissioned per hour

= 3 × 3600 × 10

^{20}/ 27.2 = 3 × 36 / 27.2 × 10^{22}= 4 × 10^{20}This gives (C) as the correct option.

askIITians offers comprehensive study material explaining the phenomenon of nuclear fission on the basis of binding energy. The JEE aspirants can refer the notes for solved examples on binding energy and solved examples on nuclear fission.

Related resources:

- Click here for the Detailed Syllabus of IIT JEE Physics.
- Look into the Sample Papers with Solutions to get a hint of the kinds of questions asked in the exam.
- You can get the knowledge of Important Books of Physics here.