If the energy of a photon corresponding to a | vedclass

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Question

(c) $E = \frac{{hc\,}}{\lambda }\,$
$ \Rightarrow \,\frac{{{E_1}}}{{{E_2}}} = \frac{{{\lambda _1}}}{{{\lambda _2}}}\,$
$\Rightarrow \,\frac{{3.32

Vedclass · Accepted Answer

$3.1$

Vedclass · Answer

(c) $E = \frac{{hc\,}}{\lambda }\,$
$ \Rightarrow \,\frac{{{E_1}}}{{{E_2}}} = \frac{{{\lambda _1}}}{{{\lambda _2}}}\,$
$\Rightarrow \,\frac{{3.32 	imes {{10}^{ - 19}}}}{{{E_2}}} = \frac{{4000}}{{6000}}$
$ \Rightarrow \,\,{E_2} = 4.98 	imes {10^{ - 19}}J = 3.1\,eV.$

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 ............ $eV$

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