In a photoelectric effect experiment,the grap | vedclass

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(C) According to Einstein's photoelectric equation,the stopping potential $V$ is given by:$eV = \frac{hc}{\lambda} - \phi$Rearranging this for $V$:$V

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Graph does not change.

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(C) According to Einstein's photoelectric equation,the stopping potential $V$ is given by:$eV = \frac{hc}{\lambda} - \phi$Rearranging this for $V$:$V = \left(\frac{hc}{e}\right)\left(\frac{1}{\lambda}\right) - \frac{\phi}{e}$Comparing this with the equation of a straight line $y = mx + c$,we have:Slope $m = \frac{hc}{e}$ and intercept $c = -\frac{\phi}{e}$.Here,$h$ is Planck's constant,$c$ is the speed of light,$e$ is the charge of an electron,and $\phi$ is the work function of the metal surface.None of these parameters $(h, c, e, \phi)$ depend on the intensity of the incident radiation. The intensity of radiation only affects the number of photoelectrons emitted per unit time (photoelectric current),not their maximum kinetic energy or the stopping potential.Therefore,changing the intensity of the incident radiation does not affect the slope or the intercept of the graph. Thus,the graph remains unchanged.

In a photoelectric effect experiment,the graph of stopping potential $V$ versus the reciprocal of wavelength $(1/\lambda)$ is shown in the figure. As the intensity of incident radiation is increased:

Similar Questions

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