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  4. The graph of kinetic energy of photoelectron versus frequency of the incident radiation is shown for two metals M and N. We may definitely conclude

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Q. The graph of kinetic energy of photoelectron versus frequency of the incident radiation is shown for two metals $M$ and $N$. We may definitely concludePhysics Question Image

COMEDKCOMEDK 2012Dual Nature of Radiation and Matter

Solution:

According to Einstein's photoelectric equation,
$E_{k} =h_{v} - \phi$
For $ M, E_{k} =0, \phi_{M } = hv_{1}$
For $ N, E_{k} = 0, \phi_{N} = hv_{2}$
$ \therefore \, \frac{\phi _{M}}{\phi _{N}} = \frac{v_{1}}{v_{2}} $
From graph,
$ v_{1 } > v_{2} $
$ \therefore \, \phi _{M} > \phi _{N }$

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