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PhysicsAtoms & Nuclei
In the first excited state of hydrogen atom, the energy of its electron is − 3·4 eV. The radial distance of the electron from the hydrogen nucleus in this case is approximately :

(Take 1 eV = 1·6 × 10⁻¹⁹ J, e = 1·6 × 10⁻¹⁹ C and 1/4πε₀ = 9 × 10⁹ N m²/C²)
Options
1
2·1 × 10⁻⁹ m
2
2·1 × 10⁻¹⁰ m
3
2·1 × 10⁻¹⁰ m
4
2·1 × 10⁻¹¹ m
Correct Answer
Option 2 : 2·1 × 10⁻¹⁰ m
Step-by-Step Solution
1

Identify the orbit number from energy:

Energy formula: \(E_n = \dfrac{-13.6}{n^2} \text{ eV}\)

Given E = −3.4 eV:

$$n^2 = \frac{13.6}{3.4} = 4 \implies n = 2$$

So the electron is in the 2nd orbit (first excited state ✓)

2

Apply Bohr's radius formula:

$$r_n = n^2 \times a_0 = n^2 \times 0.529 \text{ Å}$$

For n = 2:

$$r_2 = 4 \times 0.529 \text{ Å} = 2.116 \text{ Å}$$

3

Convert to metres:

$$r_2 = 2.116 \text{ Å} = 2.116 \times 10^{-10} \text{ m} \approx \mathbf{2.1 \times 10^{-10} \text{ m}}$$

Answer: 2.1 × 10⁻¹⁰ m

Alternative using PE method:

PE = \(-\dfrac{e^2}{4\pi\epsilon_0 r} = 2E_n\)

So \(r = \dfrac{-e^2}{4\pi\epsilon_0 \times 2E_n}\)

\(r = \dfrac{9\times10^9 \times (1.6\times10^{-19})^2}{2 \times 3.4 \times 1.6\times10^{-19}}\)

\(= \dfrac{9\times10^9 \times 2.56\times10^{-38}}{1.088\times10^{-18}} \approx 2.12 \times 10^{-10}\) m ✓

Theory: Bohr's Model of the Hydrogen Atom
1. Bohr's Postulates

Niels Bohr proposed his atomic model in 1913 to explain the stability of atoms and the discrete spectral lines of hydrogen. His model was built on three revolutionary postulates that modified classical physics.

Postulate 1 — Stationary Orbits: Electrons revolve around the nucleus in certain special circular orbits called stationary orbits or allowed orbits. While in these orbits, the electron does not radiate energy despite being accelerated (this contradicts classical electromagnetism).

Postulate 2 — Quantisation of Angular Momentum: Only those orbits are allowed for which the angular momentum of the electron is an integral multiple of h/2π: L = mvr = nh/2π, where n = 1, 2, 3... is the principal quantum number.

Postulate 3 — Radiation Condition: An electron emits or absorbs a photon only when it jumps from one allowed orbit to another. The energy of the photon equals the difference in energies of the two orbits: hν = E_higher − E_lower.

2. Key Formulas for Hydrogen Atom

\(r_n = n^2 a_0 = n^2 \times 0.529 \text{ Å}\)

\(E_n = \dfrac{-13.6}{n^2} \text{ eV}\)

\(v_n = \dfrac{v_1}{n} = \dfrac{2.18 \times 10^6}{n} \text{ m/s}\)

\(L_n = \dfrac{nh}{2\pi}\)

3. Energy Level Diagram

The energy levels of hydrogen form a series approaching zero from below. The ground state (n=1) has energy −13.6 eV, the first excited state (n=2) has −3.4 eV, the second excited state (n=3) has −1.51 eV, and so on. As n → ∞, E → 0 (ionised state).

📌 n=1: E = −13.6 eV, r = 0.529 Å (Ground state)

📌 n=2: E = −3.4 eV, r = 2.116 Å (1st excited state)

📌 n=3: E = −1.51 eV, r = 4.76 Å (2nd excited state)

📌 n=4: E = −0.85 eV, r = 8.46 Å (3rd excited state)

📌 n=∞: E = 0 eV (Ionised — electron free)

4. Excited State Naming Convention

This is a very common source of confusion in NEET. The first excited state is n=2 (one level above ground state). The second excited state is n=3. The nth excited state corresponds to principal quantum number (n+1). So always add 1 to the excited state number to get the orbit number.

⚠️ "First excited state" means n = 2, NOT n = 1

⚠️ "Second excited state" means n = 3, NOT n = 2

⚠️ Ground state is n = 1 — it is NOT an excited state

5. Energy Relations in nth Orbit

For an electron in the nth orbit of hydrogen:

📌 Total Energy: \(E_n = -\dfrac{13.6}{n^2}\) eV (always negative)

📌 Kinetic Energy: \(KE_n = +\dfrac{13.6}{n^2}\) eV (always positive)

📌 Potential Energy: \(PE_n = -\dfrac{27.2}{n^2}\) eV (always negative)

📌 Relation: KE = −E and PE = 2E, so PE = −2KE

📌 Also: E = KE + PE → −KE + PE = E ✓

6. Spectral Series of Hydrogen

When electrons jump between energy levels, they emit photons whose wavelengths form distinct series. The Rydberg formula gives the wavelength:

\(\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)\)

📌 Lyman series: n₁=1, n₂=2,3,4... → Ultraviolet region

📌 Balmer series: n₁=2, n₂=3,4,5... → Visible region

📌 Paschen series: n₁=3, n₂=4,5,6... → Infrared region

📌 Brackett series: n₁=4 → Far infrared

📌 Pfund series: n₁=5 → Far infrared

7. Limitations of Bohr's Model

Despite its success in explaining hydrogen spectra, Bohr's model has several limitations: (1) It cannot explain the spectra of multi-electron atoms. (2) It cannot explain the fine structure (splitting) of spectral lines. (3) It cannot explain the relative intensities of spectral lines. (4) It violates the Heisenberg uncertainty principle by specifying both the position and momentum of the electron precisely. (5) It cannot explain the behaviour of atoms in magnetic fields (Zeeman effect) and electric fields (Stark effect).

The modern quantum mechanical model (Schrödinger equation, orbitals) replaced Bohr's model and correctly explains all these phenomena. However, for NEET, Bohr's model is sufficient for hydrogen-like atoms.

Frequently Asked Questions
1. What orbit does the first excited state correspond to?
First excited state = n = 2. Ground state is n = 1. First excited = one step above ground = n = 2. So energy = −13.6/4 = −3.4 eV and radius = 4 × 0.529 Å = 2.116 Å.
2. What is the Bohr radius value to remember?
a₀ = 0.529 Å = 0.529 × 10⁻¹⁰ m ≈ 53 pm. This is the radius of the ground state (n=1) of hydrogen. For nth orbit: r_n = n² × 0.529 Å.
3. How do we confirm n=2 from energy −3.4 eV?
E_n = −13.6/n². Given E = −3.4: n² = 13.6/3.4 = 4, so n = 2. This is a two-step approach: first find n from energy, then find r from n. Very common NEET approach.
4. What is the relation between radius and energy in Bohr's model?
r_n ∝ n² and E_n ∝ 1/n². So r_n ∝ 1/|E_n|. Higher energy states (less negative energy) have larger radii. As energy increases towards zero, the electron moves farther from the nucleus.
5. What is the ionisation energy of hydrogen from first excited state?
Ionisation energy from n=2 = 0 − (−3.4) = 3.4 eV. Compare: from ground state (n=1) = 0 − (−13.6) = 13.6 eV. The electron in excited state needs less energy to be freed.
6. What is kinetic and potential energy in first excited state?
For n=2: Total E = −3.4 eV. Kinetic energy = +3.4 eV (= −E). Potential energy = −6.8 eV (= 2E). Check: KE + PE = 3.4 + (−6.8) = −3.4 eV = Total E ✓
7. What is Bohr's quantisation condition?
Angular momentum L = mvr = nh/2π, where n = 1,2,3... Only orbits satisfying this condition are allowed. De Broglie later explained this: the allowed orbits are those where exactly n complete de Broglie wavelengths fit: 2πr = nλ.
8. How does Bohr's model relate to de Broglie?
Bohr's quantisation mvr = nh/2π is equivalent to 2πr = nλ (de Broglie), since λ = h/mv. This shows Bohr's model is a special case of de Broglie's wave theory applied to circular orbits. Only orbits where the electron wave is a standing wave are allowed.
9. What spectral series involves transitions to n=2?
The Balmer series involves transitions from n≥3 down to n=2. These emit visible light (and some UV). The first Balmer line (n=3→2) is the red line (Hα), second (n=4→2) is blue-green (Hβ), etc.
10. What is the velocity of electron in n=2 orbit?
v₂ = v₁/n = 2.18 × 10⁶/2 = 1.09 × 10⁶ m/s. This is about 0.36% of the speed of light — fast but not relativistic, justifying Bohr's non-relativistic treatment.
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