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(C) The energy of a photon is given by the relation $E = \frac{hc}{\lambda}$.Since $h$ and $c$ are constants,we have $E \propto \frac{1}{\lambda}$.The

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$3.1$

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(C) The energy of a photon is given by the relation $E = \frac{hc}{\lambda}$.Since $h$ and $c$ are constants,we have $E \propto \frac{1}{\lambda}$.Therefore,$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.Given $\lambda_1 = 6000 \ \mathring A$,$E_1 = 3.32 	imes 10^{-19} \ J$,and $\lambda_2 = 4000 \ \mathring A$.Substituting the values: $\frac{3.32 	imes 10^{-19}}{E_2} = \frac{4000}{6000} = \frac{2}{3}$.$E_2 = \frac{3}{2} 	imes 3.32 	imes 10^{-19} \ J = 4.98 	imes 10^{-19} \ J$.To convert energy from Joules to $eV$,we divide by $1.6 	imes 10^{-19} \ J/eV$:$E_2 = \frac{4.98 	imes 10^{-19}}{1.6 	imes 10^{-19}} \ eV \approx 3.11 \ eV$.Thus,the correct option is $C$.

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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