A) \[h(v-{{v}_{0}})\]
B) \[h/v\]
C) \[he(v-{{v}_{0}})\]
D) \[h/{{v}_{0}}\]
Correct Answer: A
Solution :
Einstein explained the phenomenon of photo- electric effect on the basis of Plancks theory. According to which the kinetic energy of photoelectrons emitted from the metal surface is E and\[\phi \]is the work function of the metal, then \[E=hv-\phi \] ...(i) where\[hv\]is the energy of the photon absorbed by the electron in the metal. If for a given metal, the threshold frequency of light be\[{{v}_{0}}\] then an amount of energy\[h{{v}_{0}}\]of the photon of light will be spent in ejecting the electron out of the metal. ie, \[\phi =h{{v}_{0}}\] ...(ii) From Eqs. (i) and (ii), we get \[E=hv-h{{v}_{0}}\] \[\Rightarrow \] \[E=h(v-{{v}_{0}})\]
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