The energy of the transition from nh = 4 to nl = 2 in a hydrogen atom, expressed as E × 10³ cm⁻¹, requires calculating the wavenumber using the Rydberg formula and converting units appropriately. This transition belongs to the Balmer series in the visible spectrum. The value of E is 205.69 when rounded to two decimal places.

✅ Final Answer: E = 205.69

Rydberg Formula

The wavenumber ν̄ (in m⁻¹) for electron transitions in hydrogen is given by:

$$ \bar{\nu} = R_H \left( \frac{1}{n_l^2} – \frac{1}{n_h^2} \right) $$where RH = 1.097 × 10⁷ m⁻¹, nl = 2, nh = 4.

Substituting values yields:

$$ \bar{\nu} = 1.097 \times 10^7 \left( \frac{1}{4} – \frac{1}{16} \right) = 1.097 \times 10^7 \times \frac{3}{16} = 2.056875 \times 10^6 \, \text{m}^{-1} $$To express in cm⁻¹ (common for spectroscopy), multiply by 100: ν̄ = 2.056875 × 10⁵ cm⁻¹.

Unit Conversion

The problem states the energy as E × 10³ cm⁻¹, so:

$$ E = \frac{\bar{\nu} \, (\text{cm}^{-1})}{10^3} = \frac{2.056875 \times 10^5}{10^3} = 205.6875 $$Rounded to two decimal places: E = 205.69. This matches the Balmer series line at approximately 486 nm.

Step-by-Step Calculation

  1. Compute 1/2² = 0.25, 1/4² = 0.0625.
  2. Difference: 0.25 – 0.0625 = 0.1875.
  3. ν̄ (m⁻¹) = 1.097 × 10⁷ × 0.1875 = 2.056875 × 10⁶.
  4. Convert: 2.056875 × 10⁶ × 100 = 2.056875 × 10⁵ cm⁻¹.
  5. E = 205.69.

Balmer Series Context

The Balmer series involves transitions to n=2, producing visible light. The formula ν̄ = RH (1/2² – 1/nh²) directly applies here, with unit conversion from m⁻¹ to cm⁻¹ essential for spectroscopic energy reporting.

🎯 Practical Exam Tips (CSIR NET)

  • Always verify correct n values
  • Remember 3/16 fraction for this specific transition
  • ×100 for cm⁻¹ conversion
  • Divide by 10³ to get E value
  • Common errors: forgetting unit conversion or sign errors