The energy of a photon with a wavelength of 4 | vedclass

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Question

(A) The energy of a photon is given by the formula $E = hf = \frac{hc}{\lambda}$.Given:Planck's constant $h = 6.63 	imes 10^{-34} \, J \cdot s$Speed

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The energy of a photon is given by the formula $E = hf = \frac{hc}{\lambda}$.Given:Planck's constant $h = 6.63 	imes 10^{-34} \, J \cdot s$Speed of light $c = 3 	imes 10^8 \, m/s$Wavelength $\lambda = 450 \, nm = 450 	imes 10^{-9} \, m$Substituting the values:$E = \frac{(6.63 	imes 10^{-34}) 	imes (3 	imes 10^8)}{450 	imes 10^{-9}}$$E = \frac{19.89 	imes 10^{-26}}{450 	imes 10^{-9}}$$E = 0.0442 	imes 10^{-17} \, J = 4.42 	imes 10^{-19} \, J$Rounding to two significant figures,$E \approx 4.4 	imes 10^{-19} \, J$.

The energy of a photon with a wavelength of $450 \, nm$ is ............

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