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The light rays having photons of energy 4.2 eV are falling on a metal surface having a work function of 2.2 eV. The stopping potential of the surface is:
Detailed Solution
By Einstein's equation $h\nu=\phi+eV_s$: $4.2\ eV=2.2\ eV+eV_s \Rightarrow eV_s=2$ eV, so $V_s=2$ V.