Atomic Structure (Numericals) Solved Examples (Class 11 Chemistry)
Understanding atomic structure involves calculating energy levels, wavelengths of radiation, and other quantum mechanical properties. These examples use Bo
TL;DR: Understanding atomic structure involves calculating energy levels, wavelengths of radiation, and other quantum mechanical properties. These examples u…
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Understanding atomic structure involves calculating energy levels, wavelengths of radiation, and other quantum mechanical properties. These examples use Bo
Atomic Structure (Numericals) — Solved Numerical Examples (Step by Step)
Example 1: Calculate the energy of an electron in the 2nd Bohr orbit of a hydrogen atom. (E1 = -13.6 eV)
Solution: For hydrogen atom, the energy of an electron in the nth Bohr orbit is:
En = E1 / n²
For n = 2:
E2 = -13.6 / 2² = -13.6 / 4 = -3.4 eV
Example 2: What is the wavelength of light emitted when an electron transitions from n = 3 to n = 2 in a hydrogen atom? (R = 1.097 × 10⁷ m⁻¹)
Solution: Using Rydberg formula: 1/λ = R(1/n1² - 1/n2²)
where n1 = 2 (lower level), n2 = 3 (higher level)
1/λ = 1.097 × 10⁷ × (1/2² - 1/3²)
1/λ = 1.097 × 10⁷ × (1/4 - 1/9)
1/λ = 1.097 × 10⁷ × (9 - 4) / 36
1/λ = 1.097 × 10⁷ × 5/36
1/λ = 1.524 × 10⁶ m⁻¹
λ = 1 / (1.524 × 10⁶) = 6.56 × 10⁻⁷ m = 656 nm
Example 3: Calculate the energy required to excite an electron from n = 1 to n = 3 in hydrogen atom. (E1 = -13.6 eV)
Solution: Energy at n = 1: E1 = -13.6 eV
Energy at n = 3: E3 = -13.6 / 3² = -13.6 / 9 = -1.51 eV
Energy required = E3 - E1 = -1.51 - (-13.6) = 13.6 - 1.51 = 12.09 eV
Example 4: Find the radius of the 3rd Bohr orbit of hydrogen atom. (a0 = 0.53 Å)
Solution: For hydrogen atom, the radius of the nth Bohr orbit is:
rn = n² × a0
For n = 3:
r3 = 3² × 0.53 = 9 × 0.53 = 4.77 Å
Example 5: Calculate the frequency of light emitted in a transition from n = 4 to n = 2 in hydrogen. (c = 3 × 10⁸ m/s, R = 1.097 × 10⁷ m⁻¹)
Solution: Using Rydberg formula: 1/λ = R(1/n1² - 1/n2²)
n1 = 2, n2 = 4
1/λ = 1.097 × 10⁷ × (1/4 - 1/16)
1/λ = 1.097 × 10⁷ × (4 - 1) / 16
1/λ = 1.097 × 10⁷ × 3/16 = 2.057 × 10⁶ m⁻¹
λ = 4.86 × 10⁻⁷ m
Frequency: ν = c/λ = (3 × 10⁸) / (4.86 × 10⁻⁷)
ν = 6.17 × 10¹⁴ Hz
Example 6: What is the velocity of an electron in the 1st Bohr orbit of hydrogen? (e = 1.6 × 10⁻¹⁹ C, m = 9.11 × 10⁻³¹ kg, ε0 = 8.85 × 10⁻¹² F/m, h = 6.63 × 10⁻³⁴ J s)
Solution: For the 1st Bohr orbit (n = 1), the velocity of electron is:
v = e² / (2ε0 h × n)
Alternatively, using the relation v = c × α / n, where α = 1/137 (fine structure constant)
v1 = (3 × 10⁸) × (1/137) / 1 ≈ 2.19 × 10⁶ m/s
Tips
- Remember that energy levels in hydrogen are negative, with E increasing (becoming less negative) as n increases.
- Use the Rydberg formula for calculating wavelengths or frequencies in hydrogen transitions.
- The difference in energy between two levels equals the energy of the photon emitted or absorbed: ΔE = hν = hc/λ.
- Bohr's model works accurately only for hydrogen and hydrogen-like ions; it does not work for multi-electron atoms.
Frequently Asked Questions
Why are electron energies negative in the Bohr model?
Negative energy indicates that the electron is bound to the nucleus. The reference point (zero energy) is set at infinity, where an electron would be completely free from the atom. Any electron in an orbit is lower in energy than this free state, hence negative values.
What is the difference between emission and absorption spectra?
Emission spectrum occurs when electrons fall from higher to lower energy levels, releasing photons (light). Absorption spectrum occurs when electrons jump from lower to higher energy levels by absorbing photons. The wavelengths are the same for both, but emission appears as bright lines while absorption appears as dark lines against a bright background.
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