Quantum relation – frequency of emitted/absorbed radiation during an electronic transition For an electron transitioning between energy levels W1 and W2, the electromagnetic radiation involved has frequency f satisfying which relation?

Introduction:
Atomic and solid-state transitions emit or absorb photons whose energy equals the difference between initial and final energy levels. This fundamental quantum result, introduced by Planck and applied by Einstein and Bohr, underpins spectroscopy, lasers, LEDs, and semiconductor optoelectronics.

Given Data / Assumptions:

• Two discrete energy levels: W1 and W2.
• A single-photon process (emission or absorption).
• Planck’s constant h relates photon energy to frequency f.

Concept / Approach:

The photon energy equals the magnitude of the energy change: E_photon = |W2 − W1|. Planck’s relation states E_photon = h * f. Combining gives h * f = |W2 − W1|. The absolute value ensures positivity regardless of emission (W2 < W1) or absorption (W2 > W1).

Step-by-Step Solution:

Verification / Alternative check:

Wavelength can be found from c = f * λ; thus λ = h * c / |ΔW|, consistent with spectroscopy formulas.

Why Other Options Are Wrong:

Options B, C, D, and E rearrange or distort the relation; only a linear equality with h * f is correct.

Common Pitfalls:

Forgetting the absolute value; mixing frequency with angular frequency (ℏω) where ℏ = h / (2π).

Final Answer:

hf = |W1 - W2|