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(D) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.Since $h$ and $c$ are constants,the energy $E$ is inversely proportional t

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$4.80 	imes 10^{-19} \ J$

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(D) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.Since $h$ and $c$ are constants,the energy $E$ is inversely proportional to the wavelength $\lambda$,i.e.,$E \propto \frac{1}{\lambda}$.Therefore,we can write the ratio as $\frac{E_{2}}{E_{1}} = \frac{\lambda_{1}}{\lambda_{2}}$.Given: $E_{1} = 3.2 	imes 10^{-19} \ J$,$\lambda_{1} = 6000 \mathring{A}$,and $\lambda_{2} = 4000 \mathring{A}$.Substituting the values: $\frac{E_{2}}{3.2 	imes 10^{-19}} = \frac{6000}{4000} = \frac{6}{4} = 1.5$.$E_{2} = 1.5 	imes 3.2 	imes 10^{-19} \ J$.$E_{2} = 4.80 	imes 10^{-19} \ J$.

If the energy of a photon corresponding to a wavelength of $6000 \mathring{A}$ is $3.2 \times 10^{-19} \ J$,the photon energy for a wavelength of $4000 \mathring{A}$ will be . . . . . . .

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