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(A) According to Einstein's photoelectric equation, the maximum kinetic energy $(K_{max})$ of emitted photoelectrons is given by $K_{max} = h
u - \Ph

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(A) According to Einstein's photoelectric equation, the maximum kinetic energy $(K_{max})$ of emitted photoelectrons is given by $K_{max} = h
u - \Phi$, where $h$ is Planck's constant, $
u$ is the frequency of incident light, and $\Phi$ is the work function of the metal.This equation shows that $K_{max}$ depends only on the frequency of the incident light and the nature of the metal surface.Intensity of light affects the number of photons incident per unit area per unit time, which in turn affects the number of photoelectrons emitted (photoelectric current), but it does not affect the energy of individual photoelectrons.Therefore, when the intensity is reduced to one-fourth, the maximum kinetic energy of the emitted photoelectrons remains unchanged.

$A$ metal surface is illuminated by light of a given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum kinetic energy of the emitted photoelectrons would become

Similar Questions

The work functions for metals $A$, $B$, and $C$ are $1.92 eV$, $2.0 eV$, and $5 eV$ respectively. The metal(s) which will emit photoelectrons for incident radiation of wavelength $4100 Å$ is/are $[h=6.63 \times 10^{-34} J s, e=1.6 \times 10^{-19} C, c=3 \times 10^8 m/s]$.

The work functions for metals $A, B$,and $C$ are $1.92 \ eV, 2.0 \ eV$,and $5 \ eV$ respectively. According to Einstein's photoelectric equation,which metal$(s)$ will emit photoelectrons when irradiated with light of wavelength $4100 \ \mathring A$?

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