The energy of a photon of light with waveleng | vedclass

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(C) The energy $E$ of a photon is given by the relation $E = \frac{hc}{\lambda}$,which implies $E \propto \frac{1}{\lambda}$.Given,for $\lambda_1 = 50

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$2.5 	imes 5000 \, eV$

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(C) The energy $E$ of a photon is given by the relation $E = \frac{hc}{\lambda}$,which implies $E \propto \frac{1}{\lambda}$.Given,for $\lambda_1 = 5000 \, \mathring{A}$,$E_1 = 2.5 \, eV$.We need to find the energy $E_2$ for $\lambda_2 = 1 \, \mathring{A}$.Using the proportionality $E_1 \lambda_1 = E_2 \lambda_2$,we get:$E_2 = E_1 	imes \frac{\lambda_1}{\lambda_2}$$E_2 = 2.5 	imes \frac{5000}{1} \, eV$$E_2 = 2.5 	imes 5000 \, eV$.

The energy of a photon of light with wavelength $5000 \, \mathring{A}$ is approximately $2.5 \, eV$. In this way,the energy of an $X$-ray photon with wavelength $1 \, \mathring{A}$ would be:

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