The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = h ν (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for 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

Energy of a photon of frequency 1 Hz is given by
` E= hv`
where h= planck.s constant `= 6.63 xx 10 ^(-34) J`
` implies> E= ( 6.63 xx 10^(-34) xx 1)/( 1.6 xx 10^(-19))eV `
` = 4.14 xx 10^(-15) eV`
(a ) for radio waves , mean frequency is `3 xx 10^8 ` Hz which correspnds to photon energy of `1.24 xx 10^(-6) eV ` The source of these waves is oscillating current in LC curcuit .
(b ) for microwaves , mean frequency is `10^(10)` Hz which corresponds to photon energy of `4.14 xx 10^(-5)` eV . the source of these waves is oscillating current in specialised vacuum tubes .
(C ) for infrared radiation , mean frequency is `10^(13)` Hz which corresponds to photon energy of `4.14 xx 10 ^(-2) eV .` the source of these waves is atomic and molecular vibrations .
(d ) for visible light , the mean frequency is `6 xx 10 ^(14)` Hz which corresponds to photon energy of `2.48 ` eV . source of these waves is atomic excitation.
(e ) For ultaviolet radiations , the mean frequency is `10^(15)` Hz which corresponds to photon energy of 4.14 eV . The source of these waves is atomic excitation .
(f) for x - rays , the mean frequency is `3 xx 10^(18)` Hz which corresponds to photon energy of `1.24 xx 10^4 eV `. THe source of these waves is sudden retardation of high energy electrons by metals of high atomic numbers
(g) FOr ` gamma -`rays, the mean frequency is `3 xx 10^(20 ) `Hz which corresponds to photon energy of `1.24 xx 10^(6)` eV , The source of these waves is nuclear excitation .