Find the energy (in eV)of an optical photon of frequency 6.43 $\times$ 10¹⁴Hz. A)2.66 eV B)1.62 eV C)1.94 eV D)3.27 eV

Question

Quizplus · Accepted Answer

The energy of a photon is given by the equation E = hf, where h is Planck's constant and f is the frequency of the photon.

Plugging in the values given in the problem, we get:

E = (6.626 x 10^-34 J s) x (6.43 x 10^14 Hz)

Using the conversion factor 1 eV = 1.602 x 10^-19 J, we can convert the result from joules to electron volts:

E = (6.626 x 10^-34 J s) x (6.43 x 10^14 Hz) x (1 eV / 1.602 x 10^-19 J)

E ≈ 2.66 eV

Therefore, the correct answer is A) 2.66 eV.

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