Final answer:
The wavelength and energies per photon for the cesium lines at frequencies 3.45 × 10¹⁴ Hz and 6.53 × 10¹⁴ Hz are 870 nm (infrared, non-visible) and 460 nm (blue), and their energies are 2.286 × 10⁻¹⁹ J and 4.326 × 10⁻¹⁹ J respectively.
Step-by-step explanation:
To determine the wavelength (λ) and energy per photon of the cesium emission lines, we can use the speed of light equation c = λf, where c is the speed of light (3.00 × 108 m/s) and f is the frequency of the light. The energy per photon can be calculated using the energy equation E = hf, where h is Planck's constant (6.626 × 10-34 J·s).
a) For line (a) with the frequency of 3.45 × 1014 Hz:
- Wavelength (λ) = c/f = (3.00 × 108 m/s) / (3.45 × 1014 Hz) ≈ 8.70 × 10-7 meters or 870 nm.
- Energy (E) = hf = (6.626 × 10-34 J·s) × (3.45 × 1014 Hz) ≈ 2.286 × 10-19 joules per photon.
b) For line (b) with the frequency of 6.53 × 1014 Hz:
- Wavelength (λ) = c/f = (3.00 × 108 m/s) / (6.53 × 1014 Hz) ≈ 4.60 × 10-7 meters or 460 nm.
- Energy (E) = hf = (6.626 × 10-34 J·s) × (6.53 × 1014 Hz) ≈ 4.326 × 10-19 joules per photon.
c) The color of line (a) with a wavelength of 870 nm falls into the infrared range and therefore would not be visible to the human eye.
d) The color of line (b) corresponds to a wavelength of 460 nm, which is typically associated with a blue hue.