Given:
The wavelength of the photon is,
[tex]\begin{gathered} \lambda=560\text{ nm} \\ =560\times10^{-9}\text{ m} \end{gathered}[/tex]To find:
The frequency, energy, and the position of the photon on the electromagnetic spectrum
Explanation:
We know the speed of the electromagnetic wave in a vacuum is,
[tex]c=3\times10^8\text{ m/s}[/tex]If the frequency of the photon is 'f', we can write,
[tex]\begin{gathered} c=f\lambda \\ f=\frac{c}{\lambda} \end{gathered}[/tex]
Substituting the values we get,
[tex]\begin{gathered} f=\frac{3\times10^8}{560\times10^{-9}} \\ =5.36\times10^{14}\text{ Hz} \end{gathered}[/tex]Hence, the frequency of the photon is,
[tex]5.36\times10^{14}\text{ Hz}[/tex]The energy of the photon is,
[tex]\begin{gathered} E=hf \\ h=6.62\times10^{-34}\text{ J.s \lparen Planck's constant\rparen} \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} E=6.62\times10^{-34}\times5.36\times10^{14} \\ =3.55\times10^{-19}\text{ J} \end{gathered}[/tex]Hence, the energy of the photon is,
[tex]3.55\times10^{-19}\text{ J}[/tex]As we know the visible range is,
[tex]400\text{ nm-750 nm \lparen approx\rparen}[/tex]so, the given wavelength belongs to the visible range of the electromagnetic spectrum.
