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Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light $(v_1 > v_2)$. If the maximum values of kinetic energy of the photo electrons emitted in the two cases are in the ratio $1:n$, then the threshold frequency of the metallic surface is :

\[\begin {array} {1 1} (1)\;(v_1-v_2)/(n-1) & \quad (2)\;(nv_1-v_2)/(n-1) \\ (3)\;(nv_2-v_1)/(n-1) & \quad (4)\;(v_1-v_2)/n \end {array}\]

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1 Answer

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$ (2)\;(nv_1-v_2)/(n-1)$
answered Nov 7, 2013 by pady_1
 

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