Final answer:
The frequency of the photon emitted when an electron transitions from the 4th to the 3rd energy level is approximately 1.596 x 10^14 Hz, calculated using the energy difference and Planck’s constant.
Step-by-step explanation:
To calculate the frequency of the photon emitted when an electron transitions from the 4th energy level to the 3rd energy level in an atom, we use the formula ΔE = hf, where ΔE is the change in energy, h is Planck’s constant (6.626 × 10-34 J·s), and f is the frequency of the photon.
The change in energy (ΔE) is given by the difference in energy levels: ΔE = E3 – E4 = (-1.51 eV) – (-0.85 eV) = -0.66 eV. To convert electron volts (eV) to joules (J), we use the conversion factor 1 eV = 1.602 × 10-19 J. So, ΔE = -0.66 × 1.602 × 10-19 J/eV = -1.05732 × 10-19 J.
The negative sign indicates a release of energy, so we take the absolute value for the frequency calculation: |ΔE| = 1.05732 × 10-19 J.
ΔE = hf ⇒ f = ΔE / h = 1.05732 × 10-19 J / (6.626 × 10-34 J·s) ≈ 1.596 × 1014 Hz.
Therefore, the closest and correct answer among the options given is (b) 1.596 × 1014 Hz.