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.  

Q1. 1 atomic mass unit is equal to

(A) 1/12 (mass of one C-atom)

(B) 1/16 (mass of O₂-molecule)

(C) 1/14 (mass of N₂-moleucle)

(D) 1/25 (mass of F₂-moleucle)

Solution: We know that the atomic mass unit is defined as 1/12th of the mass of one ₆C¹² atom. Hence, here, (A) is the right answer.

Q2. 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 (U²³⁵) and plutonium (Pu) are easily fissioned by neutrons and hence (D) is the correct option. 

Q3. Energy obtained when 1 mg mass is completely converted to energy is

(A) 3×10⁸ J                            (B) 3×10¹⁰ J

(C) 9×10¹³ J                          (D) 9×10¹⁵ J

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

We know the energy is given by E = mc² 

= (1 × 10^–3) (3×10⁸)²

= 9 × 10¹³  

This gives (C) as the correct option.

Q4. 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 is

1eV = 1.6 × 10¹⁹ J

= 1.6 × 10^–13 J. Hence, this is the same as (C). 

Q5. 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. 

Q6. If 1 gm hydrogen is converted into 0.993 gm of helium, in a thermonuclear reaction, the energy released in the reaction is

(A) 63×10⁷ J                            (B) 63×10¹⁰ J

(C) 63×10¹⁴ J                           (D) 63×10²⁰ J

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

Energy E = Δmc² = (m₁ – m₂) c²

= {(1×10^–3) – (0.993×10^–3)} × (3 × 10⁸)²

= (7 × 10^–6) × (9 × 10¹⁶) = 63 × 10¹⁰ J.

Hence, (B) is the correct option.

Q7. 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.

Q8. 1 a.m.u. is equal to

(A) 1 gm                                       (B) 4.8×10^–10 esu

(C) 6.023×10²³ 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 (₆C¹²) and is numerically equal to = 1.66 × 10^–27kg. so the answer is (D).

Q9. Hydrogen bomb is based on the phenomenon of

(A) nuclear fission                          (B) nuclear fusion

(C) radioactive decay                      (D) none of these

Q10. 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

₁H² + ₁H² ——> ₁H³ + ₁H₁ + Q

Hence, in this reaction besides a tritium we get a proton. So the answer is (B). 

Q11. 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).

Q12. An atomic power reactor furnace can deliver 300 MW. The energy released due to fission of each of uranium atom U²³⁸ is 170 MeV. The number of uranium atoms fissioned per hour will be

(A) 5×10¹⁵                                          (B) 10×10²⁰

(C) 40×10²¹                                         (D) 30×10²⁵

Solution: We use the formulae for calculation of fissions per second and per hour.

The number of atoms fissioned per second is

= 3 × 10⁸ / 27.2 × 10^–12 = 3× 10²⁰ / 27.2 

No. of Atoms fissioned per hour

= 3 × 3600 × 10²⁰ / 27.2 = 3 × 36 / 27.2 × 10²² = 4 × 10²⁰

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.
 

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