Question

What is the energy of a photon of ultraviolet radiation with a frequency of 4.4 × 1015 Hz? Planck’s constant is 6.63 × 10–34

Answer 1

Answer:

A. 2.9 × 10–18 J

Explanation:

Answer 2

Answer:

$2.9\cdot 10^{-18} J$

Explanation:

The energy of a photon is given by:

$E=hf$

where

h is the Planck constant

f is the frequency of the photon

In this problem, we have:

$h = 6.63\cdot 10^{-34} Js$

$f=4.4\cdot 10^{15} Hz$ is the frequency of the photon

Substituting into the equation, we find:

$E=(6.63\cdot 10^{-34} Js)(4.4\cdot 10^{15} Hz)=2.9\cdot 10^{-18} J$