The photon energy in units of eV for an elect | vedclass

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Question

(D) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.Here,$h = 6.6 	imes 10^{-34} \, J \cdot s$,$c = 3 	imes 10^{8} \, m/s$,a

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$6.2 	imes 10^{-5}$

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(D) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.Here,$h = 6.6 	imes 10^{-34} \, J \cdot s$,$c = 3 	imes 10^{8} \, m/s$,and $\lambda = 2 \, cm = 2 	imes 10^{-2} \, m$.Substituting the values:$E = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^{8}}{2 	imes 10^{-2}} = 9.9 	imes 10^{-24} \, J$.To convert this energy into $eV$,divide by the charge of an electron $(1.6 	imes 10^{-19} \, C)$:$E = \frac{9.9 	imes 10^{-24}}{1.6 	imes 10^{-19}} \, eV = 6.1875 	imes 10^{-5} \, eV \approx 6.2 	imes 10^{-5} \, eV$.

The photon energy in units of $eV$ for an electromagnetic wave of wavelength $2 \, cm$ is

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