Question

Rank these photons in terms of decreasing energy: (b) microwave (v=9.8 ×10¹¹s⁻¹)

Answer 1

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

= Ultraviolet waves > infrared waves > microwaves

$E = hv$

where,

E = energy of photon

$v$ = frequency of the radiation

h = Planck's constant = $6.63$×$10$⁻³⁴ $J s$

We have :

(a) Frequency of infrared waves = $v$₁ = $6.5$×$10$¹³$J s$

(b) Frequency of microwaves= $v$₂ = $9.8$×$10$¹¹ $J s$

(c) Frequency of ultraviolet waves = $v$₃ = $8$×$10$¹⁵$J s$

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

$v$₃ $>$$v$₁$> v$₂

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

$E$∝$v$

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

Also the complete question is :

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)

To learn more about microwaves visit the link:

brainly.com/question/19811153

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