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# Initially a photon of wavelength $\lambda _1$ falls on photocathode and emits an electron of maximum energy $E_1$. If the wave length of the incident photon i changed to $\lambda _2$, the maximum energy of the electron emitted becomes $E_2$. Then value of hc (h= Planck's constant , c= velocity of light) is

$(a)\;hc= \frac{(E_1+E_2)\; \lambda _1 \lambda_2}{\lambda _2-\lambda _1} \quad (b)\;hc= \frac{(E_1-E_2)}{\lambda _2 - \lambda_1}.(\lambda _1. \lambda _2) \quad (c)\;hc= \frac{(E_1-E_2)\;( \lambda _2- \lambda_1)}{\lambda _1 \lambda _2} \quad (d)\;hc= \frac{\lambda _2 -\lambda_1}{\lambda _1\lambda _2E_2}.E_1$

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## 1 Answer

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$(b)\;hc= \frac{(E_1-E_2)}{\lambda _2 - \lambda_1}.(\lambda _1. \lambda _2)$
answered Nov 7, 2013 by

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