10th Class Physics Notes 2026 — Chapter 20
Chapter 20: Atomic and Nuclear Physics
Class: 10th
Subject: Physics
Board: Punjab
Chapter: 20
Exercise Solutions
A. Multiple Choice Questions (MCQs)
20.1 What did Rutherford’s alpha-particle scattering experiment reveal about the structure of the atom?
(a) Electrons are distributed evenly throughout the atom
(b) Protons and electrons are located in the nucleus
(c) Atoms are indivisible and indestructible
(d) The nucleus is small, dense, and positively charged
Correct Answer: (d) The nucleus is small, dense, and positively charged.
20.2 The atomic number of an element represents:
(a) Number of neutrons in the nucleus
(b) Number of protons in the nucleus
(c) Total number of protons and neutrons
(d) Number of electrons and neutrons
Correct Answer: (b) Number of protons in the nucleus.
20.3 Which type of radiation is positively charged and consists of two protons and two neutrons?
(a) Alpha (\(\alpha\))
(b) Beta (\(\beta\))
(c) Gamma (\(\gamma\))
(d) Neutron emission
Correct Answer: (a) Alpha (\(\alpha\)) radiation.
20.4 The process by which an unstable nucleus emits radiation and transforms into a more stable nucleus is called:
(a) Nuclear fusion
(b) Nuclear fission
(c) Radioactive decay
(d) Isotopic exchange
Correct Answer: (c) Radioactive decay.
20.5 Nuclear fission involves:
(a) Combining light nuclei to form heavier ones
(b) Splitting heavy atomic nuclei into smaller ones
(c) Absorbing gamma radiation to increase stability
(d) None of the above
Correct Answer: (b) Splitting heavy atomic nuclei into smaller ones.
20.6 What does Einstein’s equation, \(E=mc^2\), explain?
(a) The energy released in chemical reactions
(b) The equivalence of mass and energy
(c) The rate of radioactive decay
(d) The conversion of mass into momentum
Correct Answer: (b) The equivalence of mass and energy.
20.7 Which of the following is a practical application of nuclear radiation?
(a) Cooking food in microwave ovens
(b) Medical imaging and cancer treatment
(c) Solar energy generation
(d) Airplane navigation systems
Correct Answer: (b) Medical imaging and cancer treatment.
20.8 What are the potential effects of ionizing nuclear radiation on biological tissues?
(a) Increased cellular regeneration
(b) DNA damage and cancer
(c) Enhanced immune system response
(d) Improved metabolic activity
Correct Answer: (b) DNA damage and cancer.
B. Short Answer Questions
20.1 Define the atomic nucleus. What are its two main characteristics?
Answer:
The atomic nucleus is the very dense, central core of an atom that contains protons and neutrons.
Its two main characteristics are:
It contains nearly the entire mass of the atom.
It has a positive net electrical charge due to the presence of protons.
20.2 What is the structure of an atom according to Rutherford’s model?
Answer:
According to Rutherford’s nuclear model, an atom consists of a tiny, dense, positively charged central nucleus containing protons. Negatively charged electrons move around the nucleus at high speeds, and most of the atom’s volume is empty space.
20.3 How does alpha radiation differ from beta radiation?
Answer:
Alpha radiation (\(\alpha\)): It consists of helium nuclei \(\left({}^{4}_{2}\mathrm{He}\right)\) with a charge of \(+2e\). It has high ionizing power but low penetrating power and can be stopped by a sheet of paper.
Beta radiation (\(\beta\)): It consists of fast-moving, high-energy electrons \(\left({}^{0}_{-1}e\right)\). It has moderate penetrating power and can be stopped by a few millimetres of aluminium.
20.4 What is the primary energy source of the Sun and other stars?
Answer:
The primary energy source of the Sun and other stars is nuclear fusion. In the Sun, hydrogen nuclei fuse to form helium through the proton–proton chain reaction, releasing vast amounts of energy due to the mass defect, according to Einstein’s equation \(E=mc^2\).
20.5 Name the equation proposed by Einstein that relates energy and mass.
Answer:
Einstein’s mass–energy equivalence equation is:
Where:
\(E\) = Energy
\(m\) = Mass
\(c\) = Speed of light in a vacuum
20.6 What is the principle behind carbon dating, and what are its applications?
Answer:
Principle: Living organisms maintain a relatively stable ratio of radioactive carbon-14 (\(^{14}\mathrm{C}\)) to carbon-12 (\(^{12}\mathrm{C}\)) while alive. After death, the organism stops absorbing carbon, and the carbon-14 present in its remains gradually decays. Since the half-life of carbon-14 is approximately 5,730 years, measuring the remaining carbon-14 helps determine the time elapsed since death.
Applications:
Determining the age of ancient organic materials.
Dating archaeological wood and other organic remains.
Estimating the age of historical artifacts made from organic materials.
20.7 Define nuclear fusion with an example.
Answer:
Nuclear fusion is a nuclear reaction in which two light atomic nuclei combine to form a heavier nucleus, releasing a large amount of energy.
Example: Deuterium and tritium fuse to form helium and a neutron:
20.8 Americium-241 \(\left({}^{241}_{95}\mathrm{Am}\right)\) undergoes alpha decay and produces an alpha particle. Write the nuclear equation and name the nucleus formed.
Answer:
Nuclear equation:
Nucleus formed: Neptunium-237 \(\left({}^{237}_{93}\mathrm{Np}\right)\).
C. Constructed Response Questions
20.1 If an atom is mostly empty space, why do we not fall through solid objects or pass through walls?
Answer:
Although atoms are mostly empty space, their outer regions contain negatively charged electron clouds. When we touch or approach a solid object, electrostatic interactions between the electrons of our body’s atoms and those of the object’s atoms resist their overlap. These interactions, together with quantum effects, prevent ordinary matter from passing through solid objects.
20.2 Imagine a new radioactive element is discovered. How could scientists estimate its half-life without waiting thousands of years?
Answer:
Scientists can measure the radioactive decay rate by recording the activity of the sample over a short, precise time interval using a Geiger–Müller counter or another suitable detector.
The activity is related to the decay constant by:
Where:
\(A\) = Activity of the sample
\(\lambda\) = Decay constant
\(N\) = Number of undecayed nuclei
Once the decay constant is determined, the half-life can be calculated using:
20.3 If gamma rays do not change the number of protons or neutrons in a nucleus, why are they still considered dangerous?
Answer:
Gamma rays are high-frequency, high-energy electromagnetic waves. Although gamma emission does not change the atomic number or mass number of a nucleus, gamma rays can penetrate deeply into human tissue and ionize atoms and molecules. This can damage cells and DNA, potentially causing mutations and cancer.
20.4 If carbon-14 stopped forming in the atmosphere today, how would that affect the use of carbon dating in the future?
Answer:
Carbon dating depends on knowing the carbon-14 to carbon-12 ratio in the environment when an organism was alive. If carbon-14 stopped forming in the atmosphere, its atmospheric concentration would gradually decrease through radioactive decay. Establishing the original ratio for organisms living in later periods would become increasingly difficult, reducing the reliability of carbon dating unless the changing atmospheric ratio could be independently determined.
20.5 What is the difference between nuclear fission and nuclear fusion in terms of process and energy release?
Answer:
Nuclear fission: It is the process in which a heavy nucleus, such as uranium-235, splits into smaller nuclei, usually after absorbing a neutron. Energy is released because the products have a greater total binding energy than the original nucleus.
Nuclear fusion: It is the process in which two light nuclei combine to form a heavier nucleus under suitable conditions, usually requiring extremely high temperatures and pressures. Fusion can release more energy per unit mass than fission because of the increase in binding energy per nucleon.
D. Comprehensive Questions
20.1 How does the structure of an atom determine its chemical properties? Explain with reference to the arrangement of protons, neutrons, and electrons.
Answer:
The chemical properties of an atom depend mainly on its electronic configuration, particularly the arrangement of electrons in its outermost shell.
Protons: The number of protons determines the atomic number and identifies the element. In a neutral atom, the number of electrons equals the number of protons.
Electrons: Their arrangement, especially the valence electrons, determines how an atom bonds and reacts chemically with other atoms.
Neutrons: They contribute to the mass and stability of the nucleus. Different numbers of neutrons produce isotopes of the same element, generally without significantly changing its chemical properties.
20.2 Explain Einstein’s equation, \(E=mc^2\), and its significance. How does this equation relate to nuclear reactions?
Answer:
Einstein’s mass–energy equivalence equation is:
Where:
\(E\) = Energy
\(m\) = Mass
\(c\) = Speed of light in a vacuum
Significance: The equation states that mass and energy are equivalent. A small amount of mass corresponds to a very large amount of energy because the speed of light is squared.
Relation to nuclear reactions: During nuclear fission and fusion, the total mass of the products can be less than the total mass of the reactants. This difference, called the mass defect, is released as energy:
20.3 Describe the process of radioactive decay and how it leads to the emission of alpha, beta, and gamma radiation.
Answer:
Radioactive decay is the spontaneous transformation of an unstable atomic nucleus into a more stable state, accompanied by the emission of radiation.
1. Alpha decay: The nucleus emits an alpha particle \(\left({}^{4}_{2}\mathrm{He}\right)\), containing two protons and two neutrons. The mass number decreases by 4, and the atomic number decreases by 2.
2. Beta decay: In beta-minus decay, a neutron changes into a proton and emits an electron and an antineutrino. The mass number remains unchanged, while the atomic number increases by 1.
3. Gamma decay: An excited nucleus releases excess energy by emitting a high-energy electromagnetic photon called a gamma ray. The mass number and atomic number remain unchanged.
20.4 Why do different isotopes of the same element have similar chemical properties but different physical properties? Explain.
Answer:
Similar chemical properties: Isotopes of the same element have the same number of protons and, in neutral atoms, the same number of electrons. Their electronic configurations are therefore essentially the same, giving them similar chemical properties.
Different physical properties: Isotopes contain different numbers of neutrons, so their nuclear masses differ. This can affect physical properties such as mass and density, as well as nuclear stability and radioactivity.
20.5 Explain how carbon dating uses carbon-14 to estimate the age of artifacts. Discuss its reliability, considering the influence of carbon-14’s half-life.
Answer:
Mechanism: Living organisms continuously exchange carbon with their environment. While alive, their carbon-14 to carbon-12 ratio broadly reflects that of their carbon source. After death, carbon exchange stops, and the carbon-14 gradually decays. Scientists measure the remaining carbon-14 and compare it with an appropriate reference to estimate the age of the remains.
The half-life of carbon-14 is approximately 5,730 years.
Reliability and limitations: Carbon dating is useful for dating organic remains up to approximately 50,000–60,000 years old under suitable conditions. Beyond this range, the remaining carbon-14 becomes very small, making accurate measurements increasingly difficult. Contamination and changes in atmospheric carbon-14 levels can also affect the results.
E. Numerical Problems
20.1 Find the number of protons and neutrons in the nuclide \({}^{21}_{9}X\).
Given Data:
Mass number, \(A=21\)
Atomic number, \(Z=9\)
To Find:
Number of protons
Number of neutrons
Formulae:
Calculation:
1. Number of protons:
2. Number of neutrons:
Answer:
Number of protons = 9
Number of neutrons = 12
20.2 A fossil is found to have 25% of the carbon-14 compared to a living sample. How old is the fossil?
Given Data:
Remaining fraction = \(25\%=\frac14\)
Half-life of carbon-14, \(T_{1/2}=5730\) years
To Find:
Age of the fossil, \(t\).
Formula:
Since:
The sample has undergone 2 half-lives.
Calculation:
Answer: The fossil is approximately 11,460 years old.
20.3 A sample initially contains 400 g of a radioactive isotope with a half-life of 10 years. How much remains after 30 years?
Given Data:
Initial mass, \(N_0=400\text{ g}\)
Half-life, \(T_{1/2}=10\text{ years}\)
Elapsed time, \(t=30\text{ years}\)
To Find:
Remaining mass, \(N\).
Formulae:
Calculation:
1. Number of half-lives:
2. Remaining mass:
Answer: The remaining mass after 30 years is 50 g.
20.4 A radioactive substance has a half-life of 64 months. In how much time will one-eighth of the substance remain?
Given Data:
Half-life, \(T_{1/2}=64\text{ months}\)
Remaining fraction, \(\frac{N}{N_0}=\frac18\)
To Find:
Total time, \(t\).
Formulae:
Therefore, the sample undergoes 3 half-lives.
Calculation:
Answer: One-eighth of the substance will remain after 192 months.
20.5 In a nuclear reaction, \(4\,\mu\text{g}\) of mass is converted into energy. Find the energy released.
Given Data:
Mass defect, \(\Delta m=4\,\mu\text{g}\)
Speed of light, \(c=3\times10^8\text{ m/s}\)
Convert the mass into kilograms:
To Find:
Energy, \(E\).
Formula:
Calculation:
Since \(1\text{ MJ}=10^6\text{ J}\):
Answer: The energy released is \(360\text{ MJ}\).
20.6 In 420 days, one-eighth of polonium remains undecayed. Calculate the half-life of polonium.
Given Data:
Total time, \(t=420\text{ days}\)
Remaining fraction, \(\frac18\)
To Find:
Half-life, \(T_{1/2}\).
Formula:
Since:
The sample undergoes 3 half-lives. Therefore:
Calculation:
Answer: The half-life of polonium is 140 days.
20.7 Three-fourths of the initial mass of a radioactive isotope decays after one hour. Find the half-life of the isotope in minutes.
Given Data:
Fraction decayed, \(\frac34\)
Remaining fraction:
– Total time, \(t=1\text{ hour}=60\text{ minutes}\)
To Find:
Half-life, \(T_{1/2}\), in minutes.
Formula:
Therefore, the sample undergoes 2 half-lives.
Calculation:
Answer: The half-life of the isotope is 30 minutes.
Notes by Sir Dilawar
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