question_answer

Use the formula, E = hv (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?

Answer:

            Given, energy of photon, E = hv For \[~\gamma -rays\] Frequency of \[~\gamma -rays,\]  \[v=3\times {{10}^{20}}Hz\] Energy of\[~\gamma -rays,\] \[E=hv=6.6\times {{10}^{-34}}\times 3\times {{10}^{20}}\]                         \[=19.8\times {{10}^{-14}}J\] or                     \[E=\frac{19.8\times {{10}^{-14}}}{1.6\times {{10}^{-19}}}eV\]                         \[=1.24\times {{10}^{6}}eV\] The sources of \[\gamma -rays\] is nuclear decay For X-rays Frequency of X-rays,       \[~v=3\times {{10}^{18}}\text{ }Hz\] Energy of X-rays,            \[E=hv=6.6\times {{10}^{-34}}\times 3\times {{10}^{18}}\]             \[=19.8\times {{10}^{-16}}\,\,J\] or         \[E=\frac{19.8\times {{10}^{-16}}}{1.6\times {{10}^{-19}}}\]             \[=1.24\times {{10}^{4}}eV\] Retardation of high energy electrons produces X-rays. For ultraviolet rays Frequency of ultraviolet rays,        \[v={{10}^{15}}\text{ }Hz\] Energy of ultraviolet rays, \[E=hv=6.6\times {{10}^{-34}}\times {{10}^{15}}\]                         \[=6.6\times {{10}^{-19}}J\] or         \[E=\frac{6.6\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}=4.125\,eV\] It originates by the excitation of atoms. For visible rays Frequency of visible rays, \[v=6\times {{10}^{14}}\text{ }Hz\] Energy of visible rays,      \[E=hv=6.6\times {{10}^{-34}}\times 6\times {{10}^{14}}\]                         \[=39.6\times {{10}^{-20}}J\]