The energy of a photon of light with waveleng | vedclass

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(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.For the first photon:$E_1 = 2.5\,	ext{eV}$ and $\lambda_1 = 5000\,\mathri

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$2.5 	imes 5000\,	ext{eV}$

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(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.For the first photon:$E_1 = 2.5\,	ext{eV}$ and $\lambda_1 = 5000\,\mathring{A}$.So,$2.5 = \frac{hc}{5000\,\mathring{A}} \implies hc = 2.5 	imes 5000\,	ext{eV} \cdot \mathring{A}$.For the $X$-ray photon:$E_2 = \frac{hc}{\lambda_2}$ where $\lambda_2 = 1\,\mathring{A}$.Substituting the value of $hc$:$E_2 = \frac{2.5 	imes 5000}{1} = 12500\,	ext{eV}$.Thus,the energy is $2.5 	imes 5000\,	ext{eV}$.

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

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