The energy of a photon of light of wavelength | vedclass

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Question

(A) The energy $E$ of a photon is given by the formula $E = \frac{hc}{\lambda}$.Given:Planck's constant $h = 6.62 	imes 10^{-34} \ J \cdot s$Speed of

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$4.4 	imes 10^{-19} \ J$

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(A) The energy $E$ of a photon is given by the formula $E = \frac{hc}{\lambda}$.Given:Planck's constant $h = 6.62 	imes 10^{-34} \ J \cdot s$Speed of light $c = 3 	imes 10^8 \ m/s$Wavelength $\lambda = 450 \ nm = 450 	imes 10^{-9} \ m$Substituting these values into the formula:$E = \frac{6.62 	imes 10^{-34} 	imes 3 	imes 10^8}{450 	imes 10^{-9}}$$E = \frac{19.86 	imes 10^{-26}}{450 	imes 10^{-9}}$$E = 0.04413 	imes 10^{-17} \ J = 4.413 	imes 10^{-19} \ J$Rounding to two significant figures,we get $E \approx 4.4 	imes 10^{-19} \ J$.

The energy of a photon of light of wavelength $450 \ nm$ is

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