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(B) The energy $E$ of a photon is given by the formula $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$

Vedclass · Accepted Answer

$5000 : 1$

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(B) The energy $E$ of a photon is given by the formula $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.Since $h$ and $c$ are constants,the energy is inversely proportional to the wavelength: $E \propto \frac{1}{\lambda}$.Let $E_1$ be the energy of the $X$-ray photon with wavelength $\lambda_1 = 1 \mathring{A}$ and $E_2$ be the energy of the visible light photon with wavelength $\lambda_2 = 5000 \mathring{A}$.The ratio of the energies is $\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.Substituting the given values,we get $\frac{E_1}{E_2} = \frac{5000 \mathring{A}}{1 \mathring{A}} = 5000$.Therefore,the ratio is $5000 : 1$.

The ratio of the energy of an $X$-ray photon of wavelength $1 \mathring{A}$ to that of visible light of wavelength $5000 \mathring{A}$ is:

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The ratio of the energy of $X$-rays with a wavelength of $1 \ \mathring A$ to the energy of visible light with a wavelength of $5000 \ \mathring A$ is:

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