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11.Dual Nature of Radiation and matter
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If we express the energy of a photon in $KeV$ and the wavelength in angstroms, then energy of a photon can be calculated from the relation
A$E = 12.4\,h\nu $
B$E = 12.4\,h/\lambda $
C$E = 12.4\,/\lambda $
D$E = h\nu $
Solution
(c) Energy of photon $E = \frac{{hc}}{\lambda }$ (Joules) $ = \frac{{hc}}{{e\lambda }}(eV)$
$ \Rightarrow \,\mathop E\limits_{(eV)} = \frac{{6.6 \times {{10}^{ – 34}} \times 3 \times {{10}^8}}}{{1.6 \times {{10}^{ – 19}} \times \lambda ({ \mathring A})}} = \frac{{12375}}{{\lambda ({ \mathring A})}}$
$ \Rightarrow \,\,E(keV) = \frac{{12.37}}{{\lambda ({ \mathring A})}} \approx \frac{{12.4}}{\lambda }$
$ \Rightarrow \,\mathop E\limits_{(eV)} = \frac{{6.6 \times {{10}^{ – 34}} \times 3 \times {{10}^8}}}{{1.6 \times {{10}^{ – 19}} \times \lambda ({ \mathring A})}} = \frac{{12375}}{{\lambda ({ \mathring A})}}$
$ \Rightarrow \,\,E(keV) = \frac{{12.37}}{{\lambda ({ \mathring A})}} \approx \frac{{12.4}}{\lambda }$
Standard 12
Physics
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