☢️ Chapter 20: Atomic & Nuclear Physics – Numerical Problems
Prepared by Muhammad Tayyab, Subject Specialist Physics, Govt Christian High School Daska. Based on PECTAA 2026 syllabus (National Curriculum 2023).
📖 What's Inside: This section covers numerical problems from Chapter 20 Atomic and Nuclear Physics including half-life calculations, decay problems, mass-energy equivalence, and nuclear fission energy. Each problem is presented with Given Data, To Find, and step-by-step Solution as per the official PECTAA 2026 Physics curriculum. Perfect for Punjab Boards (Lahore, Gujranwala, Multan, etc.) and all BISE boards across Pakistan.
📚 Related Resources – Chapter 20
Atomic and Nuclear Physics covers atomic structure, isotopes, radioactivity, alpha/beta/gamma decay, nuclear fission, fusion, half-life, and carbon dating.
📐 Numerical Problems & Solutions (PECTAA 2026)
20.1 Find the number of protons and neutrons in the nuclide defined by \(_{9}^{21}\mathrm{X}\).
Find the number of protons and neutrons in the nuclide defined by \(_{9}^{21}\mathrm{X}\).
Given Data
\(Atomic\ number\)\(= Z = 9\)
\(Mass\ number\)\(= A = 21\)
To Find
\(Number\ of\ protons = Z = ?\)
\(Number\ of\ neutrons = N = ?\)
Solution
The atomic number \(Z\) tells us the number of protons in the nucleus.
\[\begin{aligned}
Z &= \text{number of protons} = 9
\end{aligned}\]
To find the number of neutrons, we use the formula:
\[\begin{aligned}
N &= A - Z \\
N &= 21 - 9 \\
\boldsymbol{N} &= \boldsymbol{12}
\end{aligned}\]
✅ Protons = 9, 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?
A fossil is found to have 25% of the carbon-14 compared to a living sample. How old is the fossil?
Given Data
\(Original\ quantity\)\(= N_o\)
\(Remaining\ quantity\)\(= N = 25\%\ of\ N_o\)
\(N\)\(= \frac{25}{100} \times N_o = \frac{1}{4} \times N_o\)
\(Half-life\ of\ C-14\)\(= T_{1/2} = 5730\ years\)
To Find
\(Age\ of\ fossil = t = ?\)
Solution
The number of half-lives \(n\) is given by:
\[\begin{aligned}
N &= \frac{1}{2^n} \times N_o \\
\frac{1}{4} \times N_o &= \frac{1}{2^n} \times N_o \\
\frac{1}{4} &= \frac{1}{2^n} \\
2^n &= 4 \\
2^n &= 2^2 \\
\Rightarrow n &= 2
\end{aligned}\]
Now
\[\begin{aligned}
n &= \frac{t}{T_{1/2}} \\
2 &= \frac{t}{5730\ years} \\
t &= 2 \times 5730\ years \\
\boldsymbol{t} &= \boldsymbol{11460\ years}
\end{aligned}\]
✅ \(t = 11460\ years\)
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?
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
\(Initial\ quantity\)\(= N_o = 400\ g\)
\(Half-life\)\(= T_{1/2} = 10\ years\)
\(Time\)\(= t = 30\ years\)
To Find
\(Remaining\ quantity = N = ?\)
Solution
\[\begin{aligned}
n &= \frac{t}{T_{1/2}} \\
n &= \frac{30}{10} \\
n &= 3
\end{aligned}\]
Now
\[\begin{aligned}
N &= \frac{1}{2^n} \times N_o \\
N &= \frac{1}{2^3} \times 400\ g \\
N &= \frac{1}{8} \times 400\ g \\
N &= \frac{400\ g}{8} \\
\boldsymbol{N} &= \boldsymbol{50\ g}
\end{aligned}\]
✅ \(N = 50\ g\)
20.4 A radioactive substance has a half-life of 64 months. In how much time, one-eighth of the substance will be left?
A radioactive substance has a half-life of 64 months. In how much time, one-eighth of the substance will be left?
Given Data
\(Half-life\)\(= T_{1/2} = 64\ months\)
\(Original\ quantity\)\(= N_o\)
\(Remaining\ quantity\)\(= N = \frac{1}{8} \times N_o\)
To Find
\(Time = t = ?\)
Solution
\[\begin{aligned}
\frac{1}{8} \times N_o &= \frac{1}{2^n} \times N_o \\
\frac{1}{8} &= \frac{1}{2^n} \\
2^n &= 8 \\
2^n &= 2^3 \\
\Rightarrow n &= 3
\end{aligned}\]
Now
\[\begin{aligned}
n &= \frac{t}{T_{1/2}} \\
3 &= \frac{t}{64\ months} \\
t &= 3 \times 64\ months \\
\boldsymbol{t} &= \boldsymbol{192\ months}
\end{aligned}\]
✅ \(t = 192\ months\)
20.5 In a nuclear reaction, 4 μg of mass is converted into energy. Find the energy.
In a nuclear reaction, 4 μg of mass is converted into energy. Find the energy.
Given Data
\(Mass\ converted\)\(= m = 4\mu g\)
\(m\)\(= 4 \times 10^{-6} \times 10^{-3} kg = 4 \times 10^{-9} kg\)
\(Speed\ of\ light\)\(= c = 3 \times 10^8\ ms^{-1}\)
To Find
\(Energy = E = ?\)
Solution
Using Einstein's equation:
\[\begin{aligned}
E &= mc^2 \\
E &= (4 \times 10^{-9})(3 \times 10^8)^2 \\
E &= (4 \times 10^{-9})(9 \times 10^{16}) \\
E &= 360000000\ J \\
E &= 360 \times 10^6\ J \\
\boldsymbol{E} &= \boldsymbol{360\ MJ}
\end{aligned}\]
✅ \(E = 360\ MJ\)
20.6 In 420 days, one-eighth of polonium (Po) remains undecayed. Calculate the half-life of polonium?
In 420 days, one-eighth of polonium (Po) remains undecayed. Calculate the half-life of polonium?
Given Data
\(Time\)\(= t = 420\ days\)
\(Original\ quantity\)\(= N_o\)
\(Remaining\ quantity\)\(= N = \frac{1}{8} \times N_o\)
To Find
\(Half-life = T_{1/2} = ?\)
Solution
\[\begin{aligned}
\frac{1}{8} \times N_o &= \frac{1}{2^n} \times N_o \\
\frac{1}{8} &= \frac{1}{2^n} \\
2^n &= 8 \\
2^n &= 2^3 \\
\Rightarrow n &= 3
\end{aligned}\]
Now
\[\begin{aligned}
n &= \frac{t}{T_{1/2}} \\
3 &= \frac{420\ days}{T_{1/2}} \\
T_{1/2} &= \frac{420\ days}{3} \\
\boldsymbol{T_{1/2}} &= \boldsymbol{140\ days}
\end{aligned}\]
✅ \(T_{1/2} = 140\ days\)
20.7 Three-fourth of the initial mass of a certain radioactive isotope decay after one hour. Find half-life of isotope in minutes?
Three-fourth of the initial mass of a certain radioactive isotope decay after one hour. Find half-life of isotope in minutes?
Given Data
\(Time\)\(= t = 1\ hour = 60\ minutes\)
\(Original\ quantity\)\(= N_o\)
\(Decayed\ quantity\)\(= \frac{3}{4} \times N_o\)
\(Remaining\ quantity\)\(= N = N_o - \frac{3}{4}N_o = \frac{1}{4} \times N_o\)
To Find
\(Half-life = T_{1/2} = ?\)
Solution
\[\begin{aligned}
\frac{1}{4} \times N_o &= \frac{1}{2^n} \times N_o \\
\frac{1}{4} &= \frac{1}{2^n} \\
2^n &= 4 \\
2^n &= 2^2 \\
\Rightarrow n &= 2
\end{aligned}\]
Now
\[\begin{aligned}
n &= \frac{t}{T_{1/2}} \\
2 &= \frac{60\ minutes}{T_{1/2}} \\
T_{1/2} &= \frac{60\ minutes}{2} \\
\boldsymbol{T_{1/2}} &= \boldsymbol{30\ minutes}
\end{aligned}\]
✅ \(T_{1/2} = 30\ minutes\)
📘 Examples with Solutions
Ex 20.1 Find the number of protons and neutrons in the nuclide defined by \(_{7}^{15}\mathrm{X}\).
Find the number of protons and neutrons in the nuclide defined by \(_{7}^{15}\mathrm{X}\).
Given Data
\(Atomic\ number\)\(= Z = 7\)
\(Mass\ number\)\(= A = 15\)
To Find
\(Number\ of\ protons = Z = ?\)
\(Number\ of\ neutrons = N = ?\)
Solution
From the symbol, we have atomic number:
\[\begin{aligned}
Z &= \text{number of protons} = 7
\end{aligned}\]
To find the number of neutrons, we use the formula:
\[\begin{aligned}
N &= A - Z \\
N &= 15 - 7 \\
\boldsymbol{N} &= \boldsymbol{8}
\end{aligned}\]
✅ Protons = 7, Neutrons = 8
Ex 20.2 Find the atomic mass and atomic number of the nuclei that come after alpha decay of Uranium (U-238).
Find the atomic mass and atomic number of the nuclei that come after alpha decay of Uranium (U-238).
Given Data
\(Parent\ nucleus\)\(= _{92}^{238}\mathrm{U}\)
\(Alpha\ particle\)\(= _{2}^{4}\mathrm{He}\)
To Find
\(Atomic\ mass = A = ?\)
\(Atomic\ number = Z = ?\)
Solution
Alpha decay nuclear reaction is given by:
\[
_{Z}^{A}\mathrm{X} \rightarrow _{Z-2}^{A-4}\mathrm{Y} + _{2}^{4}\mathrm{He} + \text{Energy}
\]
For Uranium-238, the nuclear reaction will be:
\[
_{92}^{238}\mathrm{U} \rightarrow _{90}^{234}\mathrm{Th} + _{2}^{4}\mathrm{He} + \text{Energy}
\]
Hence, the daughter nucleus formed after alpha decay of Uranium-238 is Thorium-234 (\(_{90}^{234}\mathrm{Th}\)) with:
\[\begin{aligned}
\text{Atomic mass} &= A = 234 \\
\text{Atomic number} &= Z = 90
\end{aligned}\]
✅ \(A = 234,\ Z = 90\) (Thorium-234)
Ex 20.3 Energy Released in Nuclear Fission
Calculate the energy released in the nuclear fission reaction: \(_{92}^{235}\mathrm{U} + _{0}^{1}\mathrm{n} \rightarrow _{56}^{141}\mathrm{Ba} + _{36}^{92}\mathrm{Kr} + 3_{0}^{1}\mathrm{n} + \text{Energy}\)
Given Data
\(Mass\ of\ _{92}^{235}\mathrm{U}\)\(= 235.0439\ u\)
\(Mass\ of\ neutron\)\(= 1.0087\ u\)
\(Mass\ of\ _{56}^{141}\mathrm{Ba}\)\(= 140.9144\ u\)
\(Mass\ of\ _{36}^{92}\mathrm{Kr}\)\(= 91.9262\ u\)
\(Mass\ of\ 3\ neutrons\)\(= 3 \times 1.0087 = 3.0261\ u\)
To Find
\(Energy\ released = E = ?\)
Solution
Mass of reactants:
\[\begin{aligned}
\text{Mass of reactants} &= \text{Mass of } _{92}^{235}\mathrm{U} + \text{Mass of neutron} \\
&= 235.0439\ u + 1.0087\ u \\
&= 236.0526\ u
\end{aligned}\]
Mass of products:
\[\begin{aligned}
\text{Mass of products} &= \text{Mass of } _{56}^{141}\mathrm{Ba} + \text{Mass of } _{36}^{92}\mathrm{Kr} + \text{Mass of 3 neutrons} \\
&= 140.9144 + 91.9262 + 3.0261 \\
&= 235.8667\ u
\end{aligned}\]
Mass defect = (Mass of reactants) - (Mass of products):
\[\begin{aligned}
\Delta m &= 236.0526 - 235.8667 \\
\Delta m &= 0.1859\ u
\end{aligned}\]
Now calculate the energy released:
\[\begin{aligned}
E &= \Delta m \times 931.5\ \frac{\text{MeV}}{u} \\
E &= 0.1859\ u \times 931.5\ \frac{\text{MeV}}{u} \\
\boldsymbol{E} &= \boldsymbol{173.2\ MeV}
\end{aligned}\]
📝 Note: Derivation of 931.5 MeV/u:
1 atomic mass unit (u) = \(1.66054 \times 10^{-27}\) kg
Speed of light = \(2.998 \times 10^8\ ms^{-1}\)
Using Einstein's equation: \(E = mc^2\)
\(E = (1.66054 \times 10^{-27})(2.998 \times 10^8)^2 = 1.4924 \times 10^{-10}\) J
Converting to MeV: \(1.4924 \times 10^{-10} \times \frac{1}{1.602 \times 10^{-13}} = 931.5\) MeV
1 atomic mass unit (u) = \(1.66054 \times 10^{-27}\) kg
Speed of light = \(2.998 \times 10^8\ ms^{-1}\)
Using Einstein's equation: \(E = mc^2\)
\(E = (1.66054 \times 10^{-27})(2.998 \times 10^8)^2 = 1.4924 \times 10^{-10}\) J
Converting to MeV: \(1.4924 \times 10^{-10} \times \frac{1}{1.602 \times 10^{-13}} = 931.5\) MeV
✅ \(E = 173.2\ MeV\)
📐 Key Formulas – Atomic & Nuclear Physics
Number of half-lives: \(n = \frac{t}{T_{1/2}}\)
Remaining quantity: \(N = \frac{1}{2^n} \times N_o\)
Mass-Energy Equivalence: \(E = mc^2\)
Number of neutrons: \(N = A - Z\)
Alpha decay: \(_{Z}^{A}\mathrm{X} \rightarrow _{Z-2}^{A-4}\mathrm{Y} + _{2}^{4}\mathrm{He}\)
Unit 10Thermal Physics
Unit 11Transfer of Thermal Energy
Unit 12Waves
Unit 13Sound
Unit 14Light
Unit 15Electrostatics
Unit 16Electricity
Unit 17Electromagnetism
Unit 18Electromagnetic Induction & EM Waves
Unit 19Electronics
Unit 20✨ Atomic and Nuclear Physics
Unit 21Space and Environment
📖 Complete syllabus coverage for Class 10 Physics (PECTAA 2026) – Units 10 to 21
💡 Exam Tip:
For numerical problems, always write Given Data, To Find, and step-by-step Solution with formulas. Pay attention to units and conversions. These problems follow the PECTAA 2026 pattern and are prepared by Subject Specialist Muhammad Tayyab.
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus