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, \[E=hv=6.6\times {{10}^{-34}}\times 6\times {{10}^{14}}\] \[=39.6\times {{10}^{-20}}J\]
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