What is the energy (in eV) of an optical photon of frequency $6.43 \times 10^{14} \mathrm {~Hz}$ (h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J) A) 2.66 eV B) 1.62 eV C) 1.94 eV D) 3.27 eV

Question

Quizplus · Accepted Answer

We can use the following equation to find the energy of the photon:

$E = h 
u$

where $E$ is the energy of the photon, $h$ is Planck's constant, and $
u$ is the frequency of the photon.

Substituting the given values, we get:

$E = (6.626 	imes 10^{-34} \mathrm{~J \cdot s})(6.43 	imes 10^{14} \mathrm{~Hz})$

$E = 4.26 	imes 10^{-19} \mathrm{~J}$

Using the conversion factor between joules and electronvolts ($1 \mathrm{~eV} = 1.60 	imes 10^{-19} \mathrm{~J}$), we get:

$E = \dfrac{4.26 	imes 10^{-19} \mathrm{~J}}{1.60 	imes 10^{-19} \mathrm{~J/eV}}$

$E = 2.66 \mathrm{~eV}$

Therefore, the answer is choice A).

What Is the Energy (In EV) of an Optical Photon 6.43×1014 Hz6.43 \times 10^{14} \mathrm {~Hz}6.43×1014 Hz

What Is the Energy (In EV) of an Optical Photon $6.43 \times 10^{14} \mathrm {~Hz}$