The energies E_1 and E_2 of two radiations ar | vedclass

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(B) The energy of a radiation 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

$\lambda_1 = 2\lambda_2$

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(B) The energy of a radiation 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,$E \propto \frac{1}{\lambda}$. Therefore,$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$. Given $E_1 = 25 \ eV$ and $E_2 = 50 \ eV$,we have: $\frac{25}{50} = \frac{\lambda_2}{\lambda_1}$ $\frac{1}{2} = \frac{\lambda_2}{\lambda_1}$ $\lambda_1 = 2\lambda_2$.

The energies $E_1$ and $E_2$ of two radiations are $25 \ eV$ and $50 \ eV$ respectively. The relation between their wavelengths,i.e.,$\lambda_1$ and $\lambda_2$,will be:

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