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

Answer:

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

= Ultraviolet waves > infrared waves > microwaves

Explanation:

$E=h\nu =\frac{h\times c}{\lambda}$

where,

E = energy of photon

$\nu$ = frequency of the radiation

h = Planck's constant = $6.63\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of the radiation

We have :

(a) Frequency of infrared waves = $\nu _1=6.5\times 10^{13} s^{-1}$

(b) Frequency of microwaves= $\nu _2=9.8\times 10^{11} s^{-1}$

(c) Frequency of ultraviolet waves = $\nu _3=8.0\times 10^{15} s^{-1}$

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

$\nu _3> \nu _1> \nu _2$

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

$E\propto \nu$

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