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Q. A and B are two metals with threshold frequencies $ 1.8\times {{10}^{14}}Hz $ and $ 2.2\times {{10}^{14}}Hz. $ Two identical photons of energy 0.825 eV each are incident on them. Then photoelectrons are emitted by $ (Take\,h=6.6\times {{10}^{-34}}J-s) $

MGIMS WardhaMGIMS Wardha 2008

Solution:

Threshold energy of A is $ {{E}_{A}}=h{{v}_{A}} $ $ =6.6\times {{10}^{-34}}\times 1.8\times {{10}^{14}} $ $ =11.88\times {{10}^{-20}}J $ $ =\frac{11.88\times {{10}^{-20}}}{1.6\times {{10}^{-19}}}eV $ $ =0.74\text{ }eV $ Similarly, $ {{E}_{B}}=0.91\text{ }eV $ Since, the incident photons have energy greater than $ {{E}_{A}} $ but less than $ {{E}_{B}} $ . So, photoelectrons will be emitted from metal A only.