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(D) The energy $E$ of a photon is related to its wavelength $\lambda$ by the equation: $E = \frac{hc}{\lambda}$.Given values are: $h = 6.63 	imes 10^

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$4.42 	imes 10^{-18}$

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(D) The energy $E$ of a photon is related to its wavelength $\lambda$ by the equation: $E = \frac{hc}{\lambda}$.Given values are: $h = 6.63 	imes 10^{-34} \ J \ s$,$c = 3 	imes 10^8 \ m \ s^{-1}$,and $\lambda = 45 \ nm = 45 	imes 10^{-9} \ m$.Substituting these values into the equation:$E = \frac{(6.63 	imes 10^{-34} \ J \ s) 	imes (3 	imes 10^8 \ m \ s^{-1})}{45 	imes 10^{-9} \ m}$$E = \frac{19.89 	imes 10^{-26}}{45 	imes 10^{-9}} \ J$$E = 0.442 	imes 10^{-17} \ J = 4.42 	imes 10^{-18} \ J$.

Calculate the energy in joule corresponding to light of wavelength $45 \ nm$. (Planck's constant,$h = 6.63 \times 10^{-34} \ J \ s$,speed of light,$c = 3 \times 10^8 \ m \ s^{-1}$)

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