CBSE Class 12 Physics Chapter 13: Nuclei NCERT Solutions

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NCERT Solutions PDF Class 12 PDF

This resource provides detailed NCERT Solutions for Class 12 Physics, Chapter 13: Nuclei. It covers fundamental concepts related to atomic structure, isotopes, and nuclear properties. The solutions explain how to calculate the atomic mass of elements based on the abundance and masses of their isotopes, using examples like Lithium and Boron. It also delves into the concept of nuclear binding energy and its relation to nuclear stability, including calculations for binding energy per nucleon. These solutions are designed to help students understand the principles of nuclear physics and prepare effectively for their board examinations by offering clear, step-by-step explanations for each exercise.

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BoardCBSE
ClassClass 12
SubjectPhysics
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 13: Nuclei - NCERT Exercises Solutions

Chapter summary

Chapter 13, Nuclei, focuses on the structure of the atomic nucleus, isotopes, isobars, and isotones. The NCERT Solutions for this chapter provide step-by-step explanations for calculating average atomic masses using isotopic abundances and masses. It also covers concepts like nuclear force, mass defect, binding energy, and the binding energy curve, which is crucial for understanding nuclear stability and reactions like fission and fusion.

Learning outcomes

  • Understand the concept of isotopes and their role in determining atomic mass.
  • Calculate the average atomic mass of an element given the masses and abundances of its isotopes.
  • Determine the abundances of isotopes when the average atomic mass and individual isotopic masses are known.
  • Explain the concept of mass defect and its relation to nuclear binding energy.
  • Calculate the binding energy per nucleon for a given nucleus.
  • Analyze the binding energy curve to understand nuclear stability.

Topics covered

Paper topics

  • Atomic Mass
  • Isotopes
  • Abundance of Isotopes
  • Mass Defect
  • Binding Energy
  • Binding Energy per Nucleon
  • Nuclear Stability
  • Average Atomic Mass Calculation
  • Nuclear Reactions (implied by binding energy curve)

Important topics

  • Calculation of Average Atomic Mass
  • Mass Defect and Binding Energy
  • Binding Energy per Nucleon
  • Binding Energy Curve and Nuclear Stability
  • Isotopes and their Abundances

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Questions and Solutions

Question 13.1

13.1 (a) Two stable isotopes of lithium <math>{}^{6}_{3}Li</math> and <math>{}^{7}_{3}Li</math> have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

(b) Boron has two stable isotopes, <math>^{^{10}B}_{^5}</math> and <math>^{^{11}B}_{^5}</math>. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of <sup>10</sup><sub>5</sub>B and <sup>11</sup><sub>5</sub>B

Solution:

Part (a): Calculating the atomic mass of Lithium

We are given the masses and abundances of the two stable isotopes of lithium:

  • Mass of <math>^{6}_{3}Li</math> isotope (<math>m_1</math>) = 6.01512 u
  • Abundance of <math>^{6}_{3}Li</math> isotope (<math>\eta_1</math>) = 7.5% = 0.075
  • Mass of <math>^{7}_{3}Li</math> isotope (<math>m_2</math>) = 7.01600 u
  • Abundance of <math>^{7}_{3}Li</math> isotope (<math>\eta_2</math>) = 92.5% = 0.925

The atomic mass of an element is the weighted average of the masses of its isotopes. The formula is:

<math display="block"> \text{Atomic Mass} = \frac{(m_1 \times \eta_1) + (m_2 \times \eta_2)}{\eta_1 + \eta_2} </math>

Since the abundances are given in percentages, their sum is 100%. So, <math>\eta_1 + \eta_2 = 7.5\% + 92.5\% = 100\% = 1</math>.

Substituting the values:

<math display="block"> \text{Atomic Mass of Li} = (6.01512 \text{ u} \times 0.075) + (7.01600 \text{ u} \times 0.925) </math>

<math display="block"> \text{Atomic Mass of Li} = 0.451134 \text{ u} + 6.489800 \text{ u} </math>

<math display="block"> \text{Atomic Mass of Li} = 6.940934 \text{ u} </math>

Therefore, the atomic mass of lithium is approximately 6.941 u.

Part (b): Calculating the abundances of Boron isotopes

We are given:

  • Mass of <math>^{10}_{5}B</math> isotope (<math>m_1</math>) = 10.01294 u
  • Mass of <math>^{11}_{5}B</math> isotope (<math>m_2</math>) = 11.00931 u
  • Atomic mass of Boron (<math>M</math>) = 10.811 u

Let the abundance of <math>^{10}_{5}B</math> be <math>x\%</math>. Then, the abundance of <math>^{11}_{5}B</math> will be <math>(100 - x)\%</math>.

Using the formula for average atomic mass:

<math display="block"> M = \frac{(m_1 \times \eta_1) + (m_2 \times \eta_2)}{\eta_1 + \eta_2} </math>

Here, <math>\eta_1 = x</math> and <math>\eta_2 = 100 - x</math>. The sum of abundances is 100.

<math display="block"> 10.811 = \frac{(10.01294 \times x) + (11.00931 \times (100 - x))}{100} </math>

Multiply both sides by 100:

<math display="block"> 1081.1 = 10.01294x + 1100.931 - 11.00931x </math>

Rearrange the terms to solve for x:

<math display="block"> 1081.1 = 1100.931 - (11.00931 - 10.01294)x </math>

<math display="block"> 1081.1 = 1100.931 - 0.99637x </math>

<math display="block"> 0.99637x = 1100.931 - 1081.1 </math>

<math display="block"> 0.99637x = 19.831 </math>

<math display="block"> x = \frac{19.831}{0.99637} \approx 19.90\% </math>

So, the abundance of <math>^{10}_{5}B</math> is approximately 19.90%.

The abundance of <math>^{11}_{5}B</math> is:

<math display="block"> 100\% - x = 100\% - 19.90\% = 80.10\% </math>

Hence, the abundance of <math>^{10}_{5}B</math> is 19.90% and that of <math>^{11}_{5}B</math> is 80.10%.

Question 13.2

13.2 The three stable isotopes of neon: <math>{}^{\frac{20}{10}}Ne, {}^{\frac{21}{10}}Ne</math> and <math>{}^{\frac{22}{10}}Ne</math> have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

Solution:

We are given the masses and abundances of the three stable isotopes of neon:

  • Isotope 1: <math>^{20}_{10}Ne</math>, Mass (<math>m_1</math>) = 19.99 u, Abundance (<math>\eta_1</math>) = 90.51% = 0.9051
  • Isotope 2: <math>^{21}_{10}Ne</math>, Mass (<math>m_2</math>) = 20.99 u, Abundance (<math>\eta_2</math>) = 0.27% = 0.0027
  • Isotope 3: <math>^{22}_{10}Ne</math>, Mass (<math>m_3</math>) = 21.99 u, Abundance (<math>\eta_3</math>) = 9.22% = 0.0922

The average atomic mass of neon is calculated as the weighted average of the masses of its isotopes. The formula is:

<math display="block"> \text{Average Atomic Mass} = \frac{(m_1 \times \eta_1) + (m_2 \times \eta_2) + (m_3 \times \eta_3)}{\eta_1 + \eta_2 + \eta_3} </math>

The sum of the abundances is <math>90.51\% + 0.27\% + 9.22\% = 100\% = 1</math>.

Substituting the given values into the formula:

<math display="block"> \text{Average Atomic Mass of Ne} = (19.99 \text{ u} \times 0.9051) + (20.99 \text{ u} \times 0.0027) + (21.99 \text{ u} \times 0.0922) </math>

Calculate each term:

  • <math>19.99 \times 0.9051 \approx 18.09295 \text{ u}</math>
  • <math>20.99 \times 0.0027 \approx 0.05667 \text{ u}</math>
  • <math>21.99 \times 0.0922 \approx 2.02718 \text{ u}</math>

Summing these values:

<math display="block"> \text{Average Atomic Mass of Ne} \approx 18.09295 + 0.05667 + 2.02718 \text{ u} </math>

<math display="block"> \text{Average Atomic Mass of Ne} \approx 20.1768 \text{ u} </math>

Therefore, the average atomic mass of neon is approximately 20.18 u.

Common mistakes

  • Incorrectly applying the formula for average atomic mass.
  • Errors in percentage to fraction conversion for abundances.
  • Calculation mistakes in mass defect and binding energy.
  • Confusing atomic mass with mass number.
  • Misinterpreting the binding energy curve.

Revision tips

  • Practice calculating average atomic masses for various elements using the provided formulas.
  • Understand the relationship between mass defect and binding energy thoroughly.
  • Memorize the key points of the binding energy curve and its implications for nuclear stability.
  • Review the definitions of isotopes, isobars, and isotones.
  • Work through all the solved examples to build confidence.

Practice MCQs

Q1. What is the primary factor determining the atomic mass of an element with multiple stable isotopes?

Q2. If an element has two isotopes with masses m1 and m2 and abundances η1 and η2, what is the formula for its average atomic mass (M)?

Q3. What does the binding energy per nucleon represent?

Q4. Which region of the binding energy curve corresponds to the most stable nuclei?

Q5. Mass defect is defined as the difference between:

Frequently asked questions

What is the main concept covered in Chapter 13: Nuclei for Class 12 Physics?

Chapter 13 focuses on the structure of the atomic nucleus, including concepts like isotopes, mass defect, binding energy, and the binding energy curve, which explains nuclear stability.

How are the atomic masses of elements with multiple isotopes calculated?

The atomic mass is calculated as the weighted average of the masses of its isotopes, where each mass is multiplied by its fractional abundance and then summed up.

What is the significance of binding energy per nucleon?

Binding energy per nucleon is a measure of the stability of a nucleus. A higher binding energy per nucleon indicates a more stable nucleus.

How can these NCERT Solutions help in exam preparation?

These solutions provide clear, step-by-step explanations for complex problems, helping students understand the underlying physics principles and practice problem-solving for exams.

What is mass defect?

Mass defect is the difference between the sum of the masses of the individual nucleons (protons and neutrons) and the actual mass of the nucleus. This mass difference is converted into binding energy.

Important links

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