Question

Consider a photon with energy 0.95 eV.

Part A)

What is its wavelength in nanometers?

Part B)

What is its frequency in hertz?

Part C)

What type of electromagnetic radiation is it?

Answer 1

Let \(E=\) energy of the photon

$$ \begin{aligned} &E=0.95 \mathrm{eV}=0.95 \times 1.6 \times 10^{-19} \mathrm{~J} \\ &E=1.52 \times 10^{-19} \mathrm{~J} \\ &\mathrm{~h}=\text { Planck's constant }=6.63 \times 10^{-34} \mathrm{Js} \\ &c=\text { speed of light } \\ &\lambda=\text { wavelength of the photon } \\ &\nu=\text { frequency of the photon } \end{aligned} $$

a) Energy of the photon is given by

$$ \begin{aligned} &E=h c / \lambda \\ &\lambda=h c / E \\ &\lambda=6.63 \times 10^{-34} \times 3 \times 10^{8} / 1.52 \times 10^{-19} \\ &\lambda=13.08 \times 10^{-7} \mathrm{~m} \end{aligned} $$

b) Energy of photon in terms of frequency is given by

$$ \begin{aligned} &\mathrm{E}=\mathrm{h} \times \nu \\ &\nu=\mathrm{E} / \mathrm{h} \\ &\nu=1.52 \times 10^{-19} / 6.63 \times 10^{-34} \\ &\nu=0.229 \times 10^{15} \\ &\nu=2.29 \times 10^{14} \mathrm{~Hz} \end{aligned} $$

c) As the frequency of the photon is of the order of \(10^{14} \mathrm{~Hz}\), so it is an infrared radiation.