Question

One type of electromagnetic radiation has a frequency of $\mathbf{107}\mathbf{.}\mathbf{1}\mathit{M}\mathit{H}\mathit{z}$, another type has a wavelength of data-custom-editor="chemistry" $\mathbf{2}\mathbf{.}\mathbf{12}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathit{m}$, and another type of electromagnetic radiation has photons with energy equal to data-custom-editor="chemistry" $\mathbf{3}\mathbf{.}\mathbf{97}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{J}\mathbf{/}$the photon. Identify each type of electromagnetic radiation, and place them in order of increasing photon energy and increasing frequency.

Short answer

Answer

Radio waves region < visible green region < X- rays region

Step-by-step solution

Frequency and energy of a wave.

The formula to calculate frequency is

$\mathit{\lambda }\mathbf{·}\mathit{v}\mathbf{=}\mathit{c}$

Where,

$\mathit{\lambda }$denotes wavelength.

v denotes frequency.

cdenotes the speed of light.

As per Planck's hypothesis, the energy associated with a photon is given as follows:

$\mathit{E}\mathbf{=}\mathit{h}\mathbf{·}\mathit{v}$

Where,

vdenotes frequency.

hdenotes Planck's constant.

Avogadro'snumber is $\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}$photon/mol.

In electromagnetic radiation, the key radiations are gamma rays, ultraviolet and UV visible light and infrared radiation. The range of wavelength for these radiations are as follows:

Radiation Wavelength-range (m)

Gamma rays ${10}^{-12}-{10}^{-10}$

X-rays ${10}^{-10}-{10}^{-8}$

Ultraviolet ${10}^{-8}-4\times {10}^{-7}$

Visible light $4\times {10}^{-7}-7\times {10}^{-7}$

Infrared ${10}^{-6}-{10}^{-2}$

Microwaves ${10}^{-2}-1$

Radiowaves $1-{10}^4$

Calculate the wavelength, frequency and energy photon associated with each wave.

The conversion factor to convert $MHz$to $Hz$is as follows:

$\mathit{1}MHz\mathit{=}{\mathit{10}}^{\mathit{6}}Hz$

Thus, $\mathit{1}\mathit{.}\mathit{071}\mathit{\times }{\mathit{10}}^{\mathit{6}}Hz$ is converted tomas follows:

localid="1656761607305" $$\begin{gathered} \mathrm{Frequency}\mathit{=}\mathit{(}\mathit{107}\mathit{.}\mathit{1}\mathit{M}\mathit{H}\mathit{z}\mathit{)}\mathit{\left(}\frac{{\mathit{10}}^{\mathit{6}}\mathit{H}\mathit{z}}{\mathit{1}\mathit{M}\mathit{H}\mathit{z}}\mathit{\right)} \\ \mathit{=}\mathit{1}\mathit{.}\mathit{071}\mathit{\times }{\mathit{10}}^{\mathit{6}}Hzn \\ \mathit{=}\mathit{1}\mathit{.}\mathit{071}\mathit{\times }{\mathit{10}}^{\mathit{6}}{s}^{\mathit{-}\mathit{1}}n \end{gathered}$$

For the second radiation,

Value of c is $\mathit{3}\mathit{\times }{\mathit{10}}^{\mathit{8}}m\mathit{/}s$.

Value of $\lambda$is $2.12\times {10}^{-10}\mathrm{m}$.

Substitute the value in the above equation.

$$\begin{gathered} v\mathit{=}\frac{\mathit{c}}{\mathit{\lambda }} \\ \mathit{}\mathit{}\mathit{}\mathit{=}\frac{\mathit{3}\mathit{\times }{\mathit{10}}^{\mathit{8}}\mathit{m}\mathit{/}\mathit{s}}{\mathit{2}\mathit{.}\mathit{12}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{10}}\mathit{m}} \\ \mathit{}\mathit{}\mathit{}\mathit{=}\mathit{1}\mathit{.}\mathit{415}\mathit{\times }{\mathit{10}}^{\mathit{18}}{s}^{\mathit{-}\mathit{1}} \end{gathered}$$

As per Planck's hypothesis, the energy associated with a photon is given as follows:

$E\mathit{=}h\mathit{·}v$

Value of vis $\mathit{1}\mathit{.}\mathit{07}\mathit{\times }{\mathit{10}}^{\mathit{6}}{s}^{\mathit{-}\mathit{1}}$.

Value of h is $\mathit{6}\mathit{.}\mathit{626}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{34}}J\mathit{}s$.

Substitute the value in the above equation to calculate the energy for a single photon.

$$\begin{gathered} E\mathit{=}h\mathit{·}v \\ \mathit{}\mathit{}\mathit{=}\left(6.626\times {10}^{-34}Js\right)\left(1.071\times {10}^6{s}^{-1}\right) \\ \mathit{}\mathit{}\mathit{=}\mathit{7}\mathit{.}\mathit{09}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{26}}\mathit{}J\mathit{/}\mathrm{photon} \end{gathered}$$

Thus, energy per photon for first radiation is $\mathit{7}\mathit{.}\mathit{09}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{26}}J\mathit{/}$a photon and it lies in the radio waves region.

For the second radiation,

Value ofv is $\mathit{1}\mathit{.}\mathit{415}\mathit{\times }{\mathit{10}}^{\mathit{18}}{s}^{\mathit{-}\mathit{1}}$.

Value of h is $\mathit{6}\mathit{.}\mathit{626}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{34}}Js$.

Substitute the value in the above equation to calculate the energy for a single photon.

$$\begin{gathered} E\mathit{=}h\mathit{·}v \\ \mathit{}\mathit{}\mathit{=}\left(6.626\times {10}^{-34}Js\right)\left(1.415\times {10}^{18}{s}^{-1}\right) \\ \mathit{}\mathit{}\mathit{=}\mathit{9}\mathit{.}\mathit{37}\mathit{\times }{\mathit{10}}^{\mathit{16}}J\mathit{/}\mathrm{photon} \end{gathered}$$

Thus energy per photon for first radiation is $\mathit{7}\mathit{.}\mathit{09}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{26}}J\mathit{/}$a photon and it lies in X the rays region.

For third radiation, the value is $\mathit{3}\mathit{.}\mathit{97}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{19}}J\mathit{/}$a photon and this corresponds to the visible green region.

Therefore, the order that shows an increase in photon energy is as follows:

Radio waves region < visible green region <X - rays region