Two radiations having energies E_1 and E_2 as | vedclass

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(B) The energy of a radiation is given by the formula $E = \frac{hc}{\lambda}$.For the first radiation: $E_1 = \frac{hc}{\lambda_1} = 15 \ eV$ $(I)$Fo

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$\lambda_1 = 3\lambda_2$

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(B) The energy of a radiation is given by the formula $E = \frac{hc}{\lambda}$.For the first radiation: $E_1 = \frac{hc}{\lambda_1} = 15 \ eV$ $(I)$For the second radiation: $E_2 = \frac{hc}{\lambda_2} = 45 \ eV$ $(II)$Dividing equation $(I)$ by equation $(II)$:$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} = \frac{15}{45} = \frac{1}{3}$Therefore,$\frac{\lambda_2}{\lambda_1} = \frac{1}{3}$,which implies $\lambda_1 = 3\lambda_2$.

Two radiations having energies $E_1$ and $E_2$ as $15 \ eV$ and $45 \ eV$ respectively,then the relationship between $\lambda_1$ and $\lambda_2$ will be:

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