Question

What is the energy of a photon of infrared radiation with a frequency of 2.53 × 1012 Hz? Planck’s constant is

6.63 × 10–34 J • s.

A. 1.68 × 1023 J      

B. 1.68 × 1047 J      C. 1.68 × 10–21 J       D. 1.68 × 10–45 J

Answer 1

C on edge

Step-by-step explanation:

Answer 2

Answer : The energy of a photon of infrared radiation is,
`1.68* 10^(-21)J`

Solution :

Formula used :

`E=h* \\u`

where,

E = energy of a photon = ?

h = Planck's constant =
`6.63* 10^(-34)Js`

`\\u` = frequency of a photon =
`2.53* 10^(12)Hz=2.53* 10^(12)s^(-1)`

Now put all the given values in this formula, we get the energy of a photon.

`E=(6.63* 10^(-34)Js)* (2.53* 10^(12)s^(-1))`

`E=1.68* 10^(-21)J`

Therefore, the energy of a photon of infrared radiation is,
`1.68* 10^(-21)J`