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Work function of a metal surface is `phi=1.5eV`. If a light of wavelength 5000Å falls on it then the maximum K.E. of ejected electron will be-A. 1.2 eVB. 0.98 eVC. 0.45 eVD. 0 eV |
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Answer» Correct Answer - B `K.E_(max)=(hc)/lamda-phi` =(12400eVÅ)/(5000Å)-1.5eV =(2.48-1.5)eV=0.98 eV |
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