Atoms Class 12 NCERT: A Deep Dive

From Plum Pudding to the Nuclear Model: A Revolution

Before we dive into the modern view, let's appreciate the journey. Initially, J.J. Thomson proposed the 'plum pudding' model, where an atom was a sphere of positive charge with electrons embedded in it, much like plums in a pudding. It was a simple idea, but it couldn't explain key observations.

The real breakthrough came from Ernest Rutherford's alpha-particle scattering experiment. Imagine firing tiny, positively charged 'bullets' (alpha particles) at a super-thin gold foil. According to Thomson's model, they should all pass through with minor deflections. However, Rutherford observed that while most particles did pass straight through, a few were deflected by large angles, and astonishingly, about 1 in 8000 bounced right back! This was as surprising as 'firing a 15-inch shell at a piece of tissue paper and it came back and hit you.'

This led to a radical new picture: the nuclear model. Rutherford concluded that the atom is mostly empty space. All the positive charge and nearly all the mass are concentrated in a tiny, dense core called the nucleus. Electrons must be orbiting this nucleus, much like planets orbit the sun. This model successfully explained the scattering results but had its own fatal flaw: according to classical physics, an orbiting (and thus accelerating) electron should continuously radiate energy, spiral inwards, and collapse into the nucleus in a fraction of a second. The stage was set for Niels Bohr to resolve this paradox.

The Postulates of Bohr's Model

Postulate 1: The Stationary Orbit Postulate

An electron can revolve around the nucleus only in certain specific circular orbits called stationary orbits. While in these orbits, the electron does not radiate any energy, contrary to the predictions of classical electromagnetism.

Postulate 2: The Angular Momentum Quantization Postulate

The angular momentum (L) of an electron in a stationary orbit is quantized. It can only be an integral multiple of h/2π, where h is Planck's constant. Mathematically, L = mvr = n(h/2π), where 'n' is the principal quantum number (n = 1, 2, 3, ...).

Postulate 3: The Frequency Postulate

An atom emits energy (in the form of a photon) only when an electron 'jumps' from a higher energy orbit (E_i) to a lower energy orbit (E_f). The frequency (ν) of the emitted photon is given by the relation hν = E_i - E_f. Similarly, an atom absorbs energy to jump from a lower to a higher orbit.

Deriving Radius and Energy of Bohr Orbits

1. Step 1: Balance the Forces — For an electron to stay in a circular orbit, the electrostatic force of attraction (F_e) between the nucleus (charge +Ze) and the electron (charge -e) must provide the required centripetal force (F_c). F_e = F_c (1 / 4πε₀) (Ze e / r²) = mv² / r
2. Step 2: Apply Bohr's Quantization Rule — From Bohr's second postulate, we know that the angular momentum is quantized: mvr = nh / 2π. From this, we can express the velocity, v = nh / (2πmr).
3. Step 3: Solve for the Radius (r_n) — Substitute the expression for 'v' from Step 2 into the force equation from Step 1. After rearranging and solving for 'r', we get the expression for the radius of the nth orbit: r_n = (n²h²ε₀) / (πme²Z) For the first orbit of a hydrogen atom (n=1, Z=1), this gives the Bohr radius, a₀ ≈ 0.529 Å.
4. Step 4: Solve for Total Energy (E_n) — The total energy (E) is the sum of Kinetic Energy (KE = ½mv²) and Potential Energy (PE = -(1/4πε₀) Ze²/r). From the force equation in Step 1, we find KE = (1/8πε₀) Ze²/r. Thus, E = KE + PE = -KE = ½PE. Substituting the expression for r_n into the energy equation gives: E_n = -(m Z² e⁴) / (8 ε₀² n² h²) For hydrogen (Z=1), the ground state energy (n=1) is E₁ ≈ -13.6 eV. The negative sign indicates that the electron is bound to the nucleus.

Worked Examples: Spectral Lines

• Problem 1: Calculate the wavelength of the second line of the Balmer series in the hydrogen spectrum. Step 1: Identify the series and transition. The Balmer series corresponds to electron transitions ending in the n_f = 2 orbit. The 'first' line is from n_i = 3 to n_f = 2. The 'second' line is therefore from n_i = 4 to n_f = 2. Step 2: Apply the Rydberg formula. The formula is 1/λ = R (1/n_f² - 1/n_i²), where R is the Rydberg constant (≈ 1.097 × 10⁷ m⁻¹). 1/λ = 1.097 × 10⁷ (1/2² - 1/4²) Step 3: Calculate the value. 1/λ = 1.097 × 10⁷ (1/4 - 1/16) 1/λ = 1.097 × 10⁷ * (3/16) 1/λ ≈ 0.2057 × 10⁷ m⁻¹ Step 4: Find the wavelength. λ = 1 / (0.2057 × 10⁷) ≈ 4.86 × 10⁻⁷ m Final Answer: The wavelength is 486 nm. This falls in the blue-green region of the visible spectrum.
• Problem 2: Find the shortest wavelength possible in the Paschen series for a hydrogen atom. Step 1: Identify the series and condition for shortest wavelength. The Paschen series involves transitions ending at n_f = 3. The shortest wavelength corresponds to the maximum energy transition, which occurs when the electron jumps from the highest possible energy level (n_i = ∞) to n_f = 3. This is also called the 'series limit'. Step 2: Apply the Rydberg formula. 1/λ_min = R (1/n_f² - 1/n_i²) 1/λ_min = R (1/3² - 1/∞²) Step 3: Calculate the value. Since 1/∞² = 0, the equation simplifies to: 1/λ_min = R / 9 1/λ_min = (1.097 × 10⁷ m⁻¹) / 9 ≈ 0.1219 × 10⁷ m⁻¹ Step 4: Find the wavelength. λ_min = 1 / (0.1219 × 10⁷) ≈ 8.204 × 10⁻⁷ m Final Answer: The shortest wavelength in the Paschen series is approximately 820.4 nm, which is in the infrared region.

Exam Traps and Key Formulas

1. The Significance of the Negative Sign: The total energy of an electron in an orbit (E_n) is always negative. This is a crucial concept. A positive or zero total energy means the electron is no longer bound to the atom (it's ionized). Don't drop the negative sign in your calculations!

2. Energy Relationships: Remember these shortcuts, they are very useful for MCQs:

• Kinetic Energy (KE) = - Total Energy (E)
• Potential Energy (PE) = 2 * Total Energy (E)
• So, KE = -1/2 PE

3. 'First Line' vs. 'Series Limit': Pay close attention to the wording. The 'first line' of a series is the transition from the very next level (n_i = n_f + 1). The 'series limit' or 'shortest wavelength' is the transition from infinity (n_i = ∞).

4. Hydrogen-like Atoms: Bohr's model and its formulas also apply to 'hydrogen-like' ions, which have only one electron (e.g., He⁺, Li²⁺). Remember to use the correct atomic number 'Z' in your formulas (Z=2 for He⁺, Z=3 for Li²⁺).

Practice Questions with Solutions

• Q: Calculate the radius of the 2nd orbit of a singly ionized helium atom (He⁺). A: Step 1: Identify the parameters. For He⁺, the atomic number Z = 2. We need the radius of the 2nd orbit, so n = 2. Step 2: Use the formula for the radius of the nth orbit: r_n = (n²h²ε₀) / (πme²Z). A simpler way is to use the relation r_n = a₀ (n²/Z), where a₀ is the Bohr radius (0.529 Å). Step 3: Substitute the values. r₂ = 0.529 Å (2² / 2) = 0.529 Å (4 / 2) = 0.529 Å 2. Step 4: Calculate the final answer. r₂ = 1.058 Å. Final answer: The radius of the 2nd orbit of He⁺ is 1.058 Å.
• Q: What is the energy required to excite a hydrogen atom from its ground state (n=1) to the third excited state (n=4)? A: Step 1: Find the energy of the initial and final states using the formula E_n = -13.6 / n² eV for hydrogen. Initial energy (ground state, n=1): E₁ = -13.6 / 1² = -13.6 eV. Final energy (third excited state means n=4): E₄ = -13.6 / 4² = -13.6 / 16 = -0.85 eV. Step 2: The required energy (ΔE) is the difference between the final and initial energy levels. ΔE = E_final - E_initial = E₄ - E₁ Step 3: Substitute the values. ΔE = (-0.85 eV) - (-13.6 eV) = -0.85 + 13.6 = 12.75 eV. Final answer: The energy required is 12.75 eV.
• Q: Calculate the longest wavelength in the Lyman series of the hydrogen spectrum. A: Step 1: Identify the series and the condition for the longest wavelength. The Lyman series corresponds to transitions ending at n_f = 1. The longest wavelength corresponds to the smallest energy transition, which is from the next level up, i.e., n_i = 2. Step 2: Apply the Rydberg formula: 1/λ_max = R (1/n_f² - 1/n_i²). 1/λ_max = R (1/1² - 1/2²) Step 3: Calculate the value. Use R ≈ 1.097 × 10⁷ m⁻¹. 1/λ_max = 1.097 × 10⁷ (1 - 1/4) = 1.097 × 10⁷ (3/4) ≈ 0.82275 × 10⁷ m⁻¹. Step 4: Invert to find the wavelength. λ_max = 1 / (0.82275 × 10⁷) ≈ 1.215 × 10⁻⁷ m. Final answer: The longest wavelength in the Lyman series is approximately 121.5 nm.
• Q: Bohr's model is a significant improvement over Rutherford's model, but it also has limitations. State two major limitations of Bohr's atomic model. A: Step 1: Recall the successes and failures of Bohr's model. It worked perfectly for hydrogen and hydrogen-like atoms but struggled with more complex atoms. Step 2: Identify the first major limitation. The model fails to explain the spectra of multi-electron atoms. The theory could not account for the electron-electron interactions that alter the energy levels. Step 3: Identify a second major limitation. It cannot explain the fine structure of spectral lines (splitting of lines into closely spaced smaller lines). Furthermore, it couldn't explain the splitting of spectral lines in the presence of magnetic fields (Zeeman effect) or electric fields (Stark effect). Final answer: Two major limitations are: 1) It is only applicable to single-electron species (like H, He⁺) and fails for multi-electron atoms. 2) It cannot explain the Zeeman effect (splitting of spectral lines in a magnetic field) or the Stark effect (splitting in an electric field).

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