The terminology different parts of the electromagnetic spectrum is given in the text. Use the formula E = hv (for energy of a quantum of radiation : photon) and obtain the photon energy in units of eV different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation ?

Explanation

Entire electromagnetic spectrum can be divided into following six sections. Corresponding to average frequency v in each section, energy of one photon is E = hv
(Where `h=6.625xx10^(-34)Js`)
(i) For section of radio waves, `v=3xx10^(8)Hz`
`E=hv=6.625xx10^(-34)xx3xx10^(8)`
`=19.88xx10^(-26)J`
`therefore E=(19.88xx10^(-26))/(1.6xx10^(-19))eV = 12.42xx10^(-7)eV`
`=1.242xx10^(-6)eV`
Oscillating charge is the source of above radiation.
(ii) For infrared waves, `v=10^(13)` Hz. Hence,
`E=hv=(6.625xx10^(-34))(10^(-13))`
`=6.625xx10^(-21)J`
`therefore E=(6.625xx10^(-21))/(1.6xx10^(-19))eV`
`therefore E=4.141xx10^(-2)eV`
Atomic and molecular excitation are the sources of above radiation.
(iii) For visible light `v=6xx10^(14)` Hz. Hence
`E=hv=(6.625xx10^(-34))(6xx10^(14))`
`=3.975xx10^(-19)J`
`therefore E=(3.975xx10^(-19))/(1.6xx10^(-19))=2.484 eV`
Transition of valence electron in an atom is the source of above radiation.
(iv) For ultraviolet light, `v=10^(15)` Hz. Hence
`E=hv=(6.625xx10^(-34))(10^(15))`
`=6.625xx10^(-19)J`
`therefore E=(6.625xx10^(-19))/(1.6xx10^(-19))=4.141 eV`
Excited atom emits above radiation.
(v) For X - rays, `v=3xx10^(18)` Hz. Hence
`E=hv=(6.625xx10^(-34))(3xx10^(18))`
`=1.988xx10^(-15)J`
`therefore E=(1.988xx10^(-15))/(1.6xx10^(-19))=1.242xx10^(4)eV`
When highly energetic electron collides and loses all of its energy suddenly, above radiation is emitted.
(vi) For `gamma` rays, `v=3xx10^(20)` Hz. Hence
`E=hv=(6.625xx10^(-34))(3xx10^(20))`
`=1.988xx10^(-13)J`
`therefore E=(1.988xx10^(-13))/(1.6xx10^(-19))=1.242xx10^(6)eV`
Excited radioactive nucleus is the source of above radiation.