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(C) The energy of the incident photon is $E = \frac{hc}{\lambda}$.Since the work function is negligible,the kinetic energy $K$ of the emitted photoele

Vedclass · Accepted Answer

$2 mc : h$

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(C) The energy of the incident photon is $E = \frac{hc}{\lambda}$.Since the work function is negligible,the kinetic energy $K$ of the emitted photoelectron is equal to the energy of the incident photon: $K = \frac{hc}{\lambda}$.The de-Broglie wavelength $\lambda_1$ of the electron is given by $\lambda_1 = \frac{h}{p}$,where $p$ is the momentum of the electron.We know that $K = \frac{p^2}{2m}$,so $p = \sqrt{2mK}$.Substituting $K = \frac{hc}{\lambda}$,we get $p = \sqrt{2m \cdot \frac{hc}{\lambda}}$.Thus,$\lambda_1 = \frac{h}{\sqrt{\frac{2mhc}{\lambda}}}$.Squaring both sides,we get $\lambda_1^2 = \frac{h^2}{\frac{2mhc}{\lambda}} = \frac{h^2 \lambda}{2mhc} = \frac{h \lambda}{2mc}$.Rearranging the terms,we find $\frac{\lambda}{\lambda_1^2} = \frac{2mc}{h}$.Therefore,the ratio $\lambda : \lambda_1^2$ is $2mc : h$.

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