In a photoelectric experiment,the collector p | vedclass

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(B) Given: Work function $\phi = 4.5\,eV$,Wavelength $\lambda = 200\,nm$.Energy of incident photon $E = \frac{hc}{\lambda} = \frac{1240\,eV\cdot nm}{2

Vedclass · Accepted Answer

The maximum kinetic energy of the photoelectrons reaching the collector is $3.7\,eV$.

Vedclass · Answer

(B) Given: Work function $\phi = 4.5\,eV$,Wavelength $\lambda = 200\,nm$.Energy of incident photon $E = \frac{hc}{\lambda} = \frac{1240\,eV\cdot nm}{200\,nm} = 6.2\,eV$.Maximum kinetic energy of emitted photoelectrons $K_{max} = E - \phi = 6.2\,eV - 4.5\,eV = 1.7\,eV$.The collector plate is at a potential of $+2.0\,V$ relative to the emitter,meaning it accelerates the electrons.The minimum kinetic energy of electrons reaching the collector is $K_{min} = K_{initial} + qV = 0 + 2.0\,eV = 2.0\,eV$.The maximum kinetic energy of electrons reaching the collector is $K_{max}' = K_{max} + qV = 1.7\,eV + 2.0\,eV = 3.7\,eV$.Thus,option $B$ is correct.

In a photoelectric experiment,the collector plate is at $2.0\,V$ with respect to the emitter plate made of copper $(\phi = 4.5\,eV)$. The emitter is illuminated by a source of monochromatic light of wavelength $200\,nm$.

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