NIOS Class 12 Physics Chapter 27 Nuclear Fission and Fusion

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NIOS Class 12 Physics Chapter 27 Nuclear Fission and Fusion

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Nuclear Fission and Fusion

Chapter: 27

Module – VII: Atoms And Nuclei

INTEXT QUESTIONS 27.1

1. Complete the following equations of nuclear reaction.

Ans: For nuclear reactions, we apply conservation laws for mass number (A) and atomic number (Z).

Given: Mass number conservation: 19 + 1 = 16 + A

→ A = 4

Atomic number conservation: 9 + 1 = 8 + Z 

→ Z = 2

Therefore, the missing particle is ⁴He₂ (alpha particle)

¹⁹F₉ + ¹H₁ → ¹⁶O₈ + ⁴He₂

(b) ²⁷Al₁₃ + ¹n₀ → ? + ⁴He₂

Mass number conservation: 27 + 1 = A + 4

→ A = 24

Atomic number conservation: 13 + 0 = Z + 2 

→ Z = 11

Therefore, the missing nucleus is ²⁴Na₁₁

²⁷Al₁₃ + ¹n₀ → ²⁴Na₁₁ + ⁴He₂

(a) ²³⁴Th₉₀ → ²³⁴Pa₉₁ + ?

Mass number conservation: 234 = 234 + A 

→ A = 0

Atomic number conservation: 90 = 91 + Z 

→ Z = -1

Therefore, the missing particle is ⁰e₋₁ (beta particle)

²³⁴Th₉₀ → ²³⁴Pa₉₁ + ⁰e₋₁

(a) ⁶³Cu₂₉ + ²D₁ → ⁶⁴Zn₃₀ + ?

Mass number conservation: 63 + 2 = 64 + A

→ A = 1

Atomic number conservation: 29 + 1 = 30 + Z 

→ Z = 0

Therefore, the missing particle is ¹n₀ (neutron)

⁶³Cu₂₉ + ²D₁ → ⁶⁴Zn₃₀ + ¹n₀

2. Calculate the energy released in the nuclear reaction given below:

Ans: First, I need to identify the missing nucleus using conservation laws.

Mass number conservation: 10 + 2 = A + 4 

→ A = 8

Atomic number conservation: 5 + 1 = Z + 2

→ Z = 4

Therefore, the missing nucleus is ⁸Be₄, but since ⁸Be is unstable, the reaction likely produces ⁷Li₃ + ¹H₁.

Let me recalculate: 10 + 2 = 7 + 4 + 1 = 12 ✓

Atomic numbers: 5 + 1 = 3 + 2 + 1 = 6 ✓

The complete reaction is: ¹⁰B₅ + ²D₁ → ⁷Li₃ + ⁴He₂ + ¹H₁

However, looking at the code execution result, I can see the calculation was done for ¹⁰B + 

²D → ¹¹C + ¹n, so let me solve accordingly.

Given: Mass of ¹⁰B = 10.01294 u

Mass of ²D = 2.014103 u

Mass of ¹¹C = 11.011434 u (from context)

Mass of neutron = 1.008665 u

1 u = 931.5 MeV

Calculate mass defect

Mass of reactants = 10.01294 + 2.014103 

= 12.027043 u

Mass of products = 11.011434 + 1.008665 

= 12.020099 u

Mass defect (Δm) = 12.027043 – 12.020099 

= 0.006944 u

Calculate energy released

Q = Δm × 931.5 MeV/u 

= 0.006944 × 931.5 

= 6.47 MeV

The energy released Q = 6.47 MeV