If the energy of a photon with wavelength 600 | vedclass

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(C) The energy of a photon is given by $E = hf = \frac{hc}{\lambda}$.Therefore, $E \propto \frac{1}{\lambda}$.Given $\lambda_1 = 6000 \ \mathring{A}$,

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$3.1$

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(C) The energy of a photon is given by $E = hf = \frac{hc}{\lambda}$.Therefore, $E \propto \frac{1}{\lambda}$.Given $\lambda_1 = 6000 \ \mathring{A}$, $E_1 = 3.32 	imes 10^{-19} \ J$.Given $\lambda_2 = 4000 \ \mathring{A}$.Using the ratio: $\frac{E_2}{E_1} = \frac{\lambda_1}{\lambda_2} = \frac{6000}{4000} = 1.5$.Thus, $E_2 = 1.5 	imes E_1 = 1.5 	imes 3.32 	imes 10^{-19} \ J = 4.98 	imes 10^{-19} \ J$.To convert energy from Joules to $eV$, divide by $1.6 	imes 10^{-19} \ J/eV$:$E_2 (in \ eV) = \frac{4.98 	imes 10^{-19}}{1.6 	imes 10^{-19}} = 3.1125 \ eV \approx 3.1 \ eV$.

If the energy of a photon with wavelength $6000 \ \mathring{A}$ is $3.32 \times 10^{-19} \ J$, then the energy of a photon with wavelength $4000 \ \mathring{A}$ is ............. $eV$.

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