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

A. 2.9 × 10–18 J

Step-by-step explanation:

Answer 2

`2.9\cdot 10^(-18) J`

Step-by-step 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`