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(A) According to Einstein's photoelectric equation, the energy of the incident photon is equal to the work function when the kinetic energy of the emi

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$ 1.04 	imes 10^{16} $

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(A) According to Einstein's photoelectric equation, the energy of the incident photon is equal to the work function when the kinetic energy of the emitted photoelectrons is zero.$ E = h
u = \phi $Where $ \phi = 110 	imes 10^{-20} \ J $ is the work function and $ h = 6.63 	imes 10^{-34} \ J \cdot s $ is Planck's constant.The frequency $ 
u $ is given by $ 
u = \frac{\phi}{h} $.The angular frequency $ \omega $ is related to frequency by $ \omega = 2\pi
u $.Substituting the values:$ \omega = 2\pi \left( \frac{\phi}{h} ight) = \frac{2 	imes 3.14 	imes 110 	imes 10^{-20}}{6.63 	imes 10^{-34}} $$ \omega \approx 1.04 	imes 10^{16} \ rad/s $.

Light is incident on a metallic plate having a work function of $ 110 \times 10^{-20} \ J $. If the produced photoelectrons have zero kinetic energy, then the angular frequency of the incident light is . . . . . . $ rad/s $. $( h = 6.63 \times 10^{-34} \ J \cdot s )$

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