Question

Rank the following photons in terms of decreasing energy:

(a) IR (v = 6.5 x 10¹³ s⁻¹)
(b) microwave (v = 9.8 x 10¹¹ s⁻¹)
(c) UV (v = 8.0 x 10¹⁵ s⁻¹)

Answer 1

The order of the energy of the photons of given wave will be

= Ultraviolet waves > infrared waves > microwaves

Step-by-step explanation:

`E=h\\u =(h* c)/(\lambda)`

where,

E = energy of photon

`\\u` = frequency of the radiation

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

c = speed of light =
`3* 10^8m/s`

`\lambda` = wavelength of the radiation

We have :

(a) Frequency of infrared waves =
`\\u _1=6.5* 10^(13) s^(-1)`

(b) Frequency of microwaves=
`\\u _2=9.8* 10^(11) s^(-1)`

(c) Frequency of ultraviolet waves =
`\\u _3=8.0* 10^(15) s^(-1)`

So, the decreasing order of the frequencies of the waves will be :

`\\u _3> \\u _1> \\u _2`

As we can see from the formula that energy is directly proportional to the frequency of the wave.

`E\propto \\u`

So, the order of the energy of the photons of given wave will be same as their order of frequencies:

`E_3>E_1>E_2`

= Ultraviolet waves > infrared waves > microwaves