Answer:
Kinetic energy of photoelectrons is given by Einstein's photoelectric equation, \[{{K}_{\max }}=\frac{1}{2}m\upsilon _{\max }^{2}\] \[=hv-{{W}_{0}}=\frac{hc}{\lambda }-{{W}_{0}}\] \[\therefore \] \[\upsilon _{\max }^{2}\propto \frac{1}{\lambda }\] or \[{{\upsilon }_{\max }}\propto \frac{1}{\sqrt{\lambda }}\] As the wavelength of incident light decreases, the velocity of photoelectrons increases.
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