91Ó°ÊÓ

A photon is a particle-like packet of electromagnetic energy. The formula to calculate the energy (E) of a photon is equal to:

$\mathbf{E}\mathbf{=}\mathbf{³óν}\mathbf{or}\frac{\mathbf{\text{hc}}}{\mathbf{λ}}$

Where,

E is the energy of the photon

$\mathit{ν}$is the frequency of electromagnetic radiation and$\mathit{ν}\mathbf{=}\frac{\mathbf{c}}{\mathbf{λ}}$

c is the speed of light in a vacuum$\mathbf{=}\mathbf{3}\mathbf{×}{\mathbf{10}}^{\mathbf{8}}\mathbf{m}\mathbf{/}\mathbf{s}$

$\mathit{λ}$is the wavelength of electromagnetic radiation.

h is Planck's constant$\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{62}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathbf{Js}$

From the formula, it can be seen energy depends on wavelength. Energy and wavelength are inversely proportional. So, the more the wavelength, the less will be energy.

Energy is directly proportional to the frequency of radiation. So, the more the frequency more will be energy.

• From the given data, it can be seen that wavelength is more for infrared radiation than X-ray. So, X-ray has more energy than infrared radiation.
• For, $\mathrm{ν}=4.0×{10}^{9}Hz$

$$\begin{gathered} \mathrm{E}=\mathrm{³óν} \\ =6.62×{10}^{-34}×4.0×{10}^9 \\ =2.65×{10}^{-24}\mathrm{J} \end{gathered}$$

$$\begin{gathered} \mathrm{E}=\frac{\mathrm{hc}}{\mathrm{λ}} \\ =\frac{6.62×{10}^{-34}×3×{10}^8}{9×{10}^{-6}} \\ =2.21×{10}^{-20}\mathrm{J} \end{gathered}$$