A photon of frequency 6.40 × 10¹⁴ Hz has what energy? A) 1.26 eV B) 1.93 eV C) 2.11 eV D) 2.65 eV E) 3.17 eV

Question

Quizplus · Accepted Answer

The energy of a photon is given by the formula E = hf, where h is the Planck's constant (6.626 × 10^-34 J·s) and f is the frequency of the photon in Hz. 
So, E = (6.626 × 10^-34 J·s) × (6.40 × 10^14 Hz) = 4.23 × 10^-19 J 
To convert this energy into eV, we use the conversion factor: 1 eV = 1.60 × 10^-19 J 
So, E = (4.23 × 10^-19 J) / (1.60 × 10^-19 J/eV) = 2.65 eV 
Therefore, the answer is D: 2.65 eV

A Photon of Frequency 6