The wavelength (in mathring{A}) of a photon h | vedclass

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(C) The energy of the photon is given as $E = 3 \ eV$.Converting energy to $erg$: $E = 3 	imes 1.6 	imes 10^{-12} \ erg = 4.8 	imes 10^{-12} \ erg$

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$4141$

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(C) The energy of the photon is given as $E = 3 \ eV$.Converting energy to $erg$: $E = 3 	imes 1.6 	imes 10^{-12} \ erg = 4.8 	imes 10^{-12} \ erg$.Using the formula $\lambda = \frac{hc}{E}$:$\lambda = \frac{6.626 	imes 10^{-27} \ erg \ s 	imes 3 	imes 10^{10} \ cm/s}{4.8 	imes 10^{-12} \ erg}$.$\lambda = \frac{19.878 	imes 10^{-17}}{4.8 	imes 10^{-12}} \ cm = 4.141 	imes 10^{-5} \ cm$.Since $1 \ cm = 10^8 \ \mathring{A}$, then $\lambda = 4.141 	imes 10^{-5} 	imes 10^8 \ \mathring{A} = 4141 \ \mathring{A}$.

The wavelength (in $\mathring{A}$) of a photon having energy $3 \ eV$ is approximately.
$1 \ eV = 1.6 \times 10^{-12} \ erg$
$h = 6.626 \times 10^{-27} \ erg \ s$
$c = 3 \times 10^{10} \ cm/s$

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The wavelength (in $\mathring{A}$) of a photon having energy $3 \ eV$ is approximately.$1 \ eV = 1.6 \times 10^{-12} \ erg$$h = 6.626 \times 10^{-27} \ erg \ s$$c = 3 \times 10^{10} \ cm/s$