Atoms and Nuclei | CBSE - Wyatt's Notes

sources:

• text: Standard textbook reference

This topic covers the Bohr model of the hydrogen atom, X-ray production, nuclear structure, radioactivity, and nuclear energy.

• Bohr radius: $a_0 = \frac{\varepsilon_0 h^2}{\pi m e^2} \approx 0.529 \, \text{\AA}$
• Energy levels: $E_n = \frac{-13.6}{n^2} \, \text{eV}$ for hydrogen
• Rydberg formula: $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ where $R = 1.097 \times 10^7 \, \text{m}^{-1}$
• Nuclear mass defect: $\Delta m = Zm_p + Nm_n - M$
• Binding energy: $E_B = \Delta m \, c^2$
• Radioactive decay: $N = N_0 e^{-\lambda t}$, half-life $t_{1/2} = \frac{0.693}{\lambda}$
• Mass-energy equivalence: $E = mc^2$
• Alpha decay: $^A_Z X \rightarrow ^{A-4}_{Z-2} Y + ^4_2 \text{He}$
• Beta decay: $^A_Z X \rightarrow ^{A}_{Z+1} Y + e^- + \bar{\nu}_e$

Problem: Find the wavelength of light emitted when an electron in hydrogen transitions from $n = 3$ to $n = 2$.

Solution:

Energy levels: $E_3 = \frac{-13.6}{9} = -1.51 \, \text{eV}, \quad E_2 = \frac{-13.6}{4} = -3.40 \, \text{eV}$

Energy of emitted photon: $\Delta E = E_3 - E_2 = -1.51 - (-3.40) = 1.89 \, \text{eV}$

Convert to wavelength: $\lambda = \frac{hc}{\Delta E} = \frac{1240}{1.89} = 656 \, \text{nm}$

This is the red line of the Balmer series.

Common mistake: Using $E_n = -13.6/n^2$ without the negative sign. The energy is negative because the electron is bound.

Problem: Calculate the binding energy per nucleon of $^4_2\text{He}$. (Mass of proton = 1.00728 u, mass of neutron = 1.00867 u, mass of $^4$He = 4.00260 u)

Solution:

Mass defect: $\Delta m = 2(1.00728) + 2(1.00867) - 4.00260$ $= 2.01456 + 2.01734 - 4.00260 = 0.02930 \, \text{u}$

Binding energy: $E_B = \Delta m \times 931.5 \, \text{MeV/u} = 0.02930 \times 931.5 = 27.3 \, \text{MeV}$

Binding energy per nucleon: $\frac{E_B}{A} = \frac{27.3}{4} = 6.83 \, \text{MeV/nucleon}$

Common mistake: Forgetting to multiply by 931.5 to convert mass defect from atomic mass units to MeV.

Problem: A radioactive sample has a half-life of 10 days. What fraction remains after 30 days?

Solution:

Number of half-lives: $n = \frac{30}{10} = 3$

Fraction remaining: $\frac{N}{N_0} = \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^3 = \frac{1}{8} = 0.125$

Alternatively, using the decay formula: $\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{10} = 0.0693 \, \text{day}^{-1}$

$N = N_0 e^{-\lambda t} = N_0 e^{-0.0693 \times 30} = N_0 e^{-2.079} = 0.125 \, N_0$

Common mistake: Using $N = N_0 e^{-t/t_{1/2}}$ instead of $N = N_0 e^{-\lambda t}$ where $\lambda = \ln 2 / t_{1/2}$.

1. Find the wavelength of the first line of the Lyman series for hydrogen.
2. Calculate the binding energy of $^{56}_{26}\text{Fe}$ given its mass is 55.9349 u.
3. A radioactive substance decays to 1/16 of its original amount in 40 days. Find its half-life.

• Bohr model problems involve energy level transitions and spectral series
• Use $\lambda = hc / \Delta E$ to convert energy differences to wavelengths
• Binding energy problems always involve mass defect and $E = mc^2$
• Radioactive decay problems use either the half-life or the decay constant
• The Lyman series (UV), Balmer series (visible), and Paschen series (IR) correspond to different final states

• Energy levels: $E_n = \frac{-13.6}{n^2}$ eV (hydrogen)
• Rydberg formula: $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$
• de Broglie wavelength: $\lambda = \frac{h}{mv}$
• Mass-energy equivalence: $E = mc^2$, $1\,\text{u} = 931.5\,\text{MeV}/c^2$
• Radioactive decay: $N = N_0 e^{-\lambda t}$, $t_{1/2} = \frac{0.693}{\lambda}$
• Activity: $A = \lambda N = A_0 e^{-\lambda t}$

Problem: Find the de Broglie wavelength of an electron accelerated through a potential difference of 100 V.

Solution:

Kinetic energy gained: $eV = \frac{1}{2}mv^2$

Momentum: $p = mv = \sqrt{2meV}$

de Broglie wavelength: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$

$= \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times 100}}$

$= \frac{6.63 \times 10^{-34}}{\sqrt{2.915 \times 10^{-47}}} = \frac{6.63 \times 10^{-34}}{5.40 \times 10^{-24}} = 1.23 \times 10^{-10}\,\text{m} = 0.123\,\text{nm}$

Common mistake: Forgetting to convert electron-volts to joules. $1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$.

Problem: Complete the nuclear reaction: $^{238}_{92}\text{U} \rightarrow \, ? + \, ^4_2\text{He}$

Solution:

Conservation of mass number: $238 = A + 4 \implies A = 234$

Conservation of atomic number: $92 = Z + 2 \implies Z = 90$

The product is $^{234}_{90}\text{Th}$ (thorium-234).

$^{238}_{92}\text{U} \rightarrow \, ^{234}_{90}\text{Th} + \, ^4_2\text{He}$

Common mistake: Forgetting to conserve both mass number and atomic number. Both must balance on both sides.

Problem: A radioactive sample has a half-life of 5 years. If its initial activity is 800 Bq, what is the activity after 15 years?

Solution:

Number of half-lives: $n = \frac{15}{5} = 3$

Activity after $n$ half-lives: $A = A_0 \left(\frac{1}{2}\right)^n = 800 \times \left(\frac{1}{2}\right)^3 = 800 \times \frac{1}{8} = 100\,\text{Bq}$

Alternatively, using the decay formula: $\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5} = 0.1386\,\text{year}^{-1}$

$A = A_0 e^{-\lambda t} = 800 \times e^{-0.1386 \times 15} = 800 \times e^{-2.079} = 800 \times 0.125 = 100\,\text{Bq}$

Common mistake: Using the wrong formula for activity. Activity follows the same exponential decay law as the number of radioactive nuclei.

1. For Bohr model problems, remember that the electron orbits in the $n$-th orbit with radius $r_n = n^2 a_0$
2. The Rydberg formula gives wavelengths of emitted/absorbed light; use $n_1 < n_2$ for emission
3. In nuclear reactions, always conserve mass number, atomic number, and charge
4. For binding energy calculations, the mass defect is always positive (products have less mass than constituents)
5. Activity is measured in Becquerels (Bq): 1 Bq = 1 decay per second

Atoms are tiny solar systems where electrons orbit the nucleus in quantized energy levels. When an electron jumps between levels, it emits or absorbs a photon with energy exactly equal to the gap — this is why atoms produce discrete spectral lines rather than a continuous rainbow. The nucleus holds protons and neutrons together with the strong nuclear force, and the mass defect tells you how much energy is locked in that bond. Radioactive decay is random at the individual atom level but predictable in bulk: the half-life tells you how long it takes for half the atoms to decay, regardless of how many you start with.

The energy levels of hydrogen are $E_n = -13.6/n^2$ eV, and the negative sign indicates a bound state. Students often forget the negative sign when calculating transition energies, leading to incorrect wavelengths. When computing $\Delta E = E_{final} - E_{initial}$, always include the signs: $\Delta E = (-13.6/n_f^2) - (-13.6/n_i^2)$. A positive $\Delta E$ means a photon was emitted.

The mass number $A$ is the count of nucleons (an integer), while the atomic mass is the measured mass in unified atomic mass units. Students sometimes use the mass number directly in the mass defect formula $\Delta m = Zm_p + Nm_n - M$ instead of the actual atomic mass. Always use the tabulated atomic mass (e.g., 4.00260 u for helium-4), not the mass number (4).

The radioactive decay law is $N = N_0 e^{-\lambda t}$ where $\lambda = \ln 2 / t_{1/2}$. A common error is writing $N = N_0 e^{-t/t_{1/2}}$ or $N = N_0 (1/2)^{t/t_{1/2}}$ interchangeably without understanding they are equivalent only when $\lambda = \ln 2 / t_{1/2}$. The exponential form with $\lambda$ is preferred for calculus-based problems, while the half-life form is quicker for integer multiples of half-lives.

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