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11.Dual Nature of Radiation and matter
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If the energy of a photon corresponding to a wavelength of $6000 \mathring A$ is $3.32 \times {10^{ - 19}}J$, the photon energy for a wavelength of $4000 \mathring A$ will be ............ $eV$
A$1.4 $
B$4.9 $
C$3.1$
D$1.6 $
Solution
(c) $E = \frac{{hc\,}}{\lambda }\,$
$ \Rightarrow \,\frac{{{E_1}}}{{{E_2}}} = \frac{{{\lambda _1}}}{{{\lambda _2}}}\,$
$\Rightarrow \,\frac{{3.32 \times {{10}^{ – 19}}}}{{{E_2}}} = \frac{{4000}}{{6000}}$
$ \Rightarrow \,\,{E_2} = 4.98 \times {10^{ – 19}}J = 3.1\,eV.$
$ \Rightarrow \,\frac{{{E_1}}}{{{E_2}}} = \frac{{{\lambda _1}}}{{{\lambda _2}}}\,$
$\Rightarrow \,\frac{{3.32 \times {{10}^{ – 19}}}}{{{E_2}}} = \frac{{4000}}{{6000}}$
$ \Rightarrow \,\,{E_2} = 4.98 \times {10^{ – 19}}J = 3.1\,eV.$
Standard 12
Physics
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