A) 0.5 eV
B) 1.1 eV
C) 2.0 eV
D) 1.5 eV
Correct Answer: B
Solution :
If the maximum kinetic energy of photo electrons emitted from metal surface is \[[ML{{T}^{-2}}]\] and W is the work-function of metal then \[=[M{{L}^{-1}}{{T}^{-2}}]\] where hv is the energy of photon absorbed by the electron in metal. \[[M{{L}^{2}}{{T}^{-1}}]\] \[\alpha =\frac{\Delta {{i}_{C}}}{\Delta {{i}_{g}}}\] where \[\Delta {{i}_{C}}\] Putting the numerical values, we have \[\Delta {{i}_{g}}\] \[\alpha =0.96,\Delta {{i}_{E}}=7.2,\] Note: Energy of incident photons should be greater than work function of metal for emission of photo electrons to take place.
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