In an experiment on the photoelectric effect, | vedclass

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(A) The energy of the incident photons is given by $E = \frac{hc}{\lambda}$.For $\lambda_1 = 550\ nm$,$E_1 = \frac{1240}{550} \approx 2.25\ eV$.For $\

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(A) The energy of the incident photons is given by $E = \frac{hc}{\lambda}$.For $\lambda_1 = 550\ nm$,$E_1 = \frac{1240}{550} \approx 2.25\ eV$.For $\lambda_2 = 450\ nm$,$E_2 = \frac{1240}{450} \approx 2.75\ eV$.For $\lambda_3 = 350\ nm$,$E_3 = \frac{1240}{350} \approx 3.54\ eV$.For metal $r$ $(\phi_r = 3.0\ eV)$,only $\lambda_3$ causes emission $(E_3 > \phi_r)$.For metal $q$ $(\phi_q = 2.5\ eV)$,$\lambda_2$ and $\lambda_3$ cause emission $(E_2, E_3 > \phi_q)$.For metal $p$ $(\phi_p = 2.0\ eV)$,all three $\lambda_1, \lambda_2, \lambda_3$ cause emission $(E_1, E_2, E_3 > \phi_p)$.Since the intensity is the same,the saturation current $I$ depends on the number of photons incident and the number of wavelengths that can cause emission. Metal $p$ allows emission for all three wavelengths,resulting in the highest saturation current,while metal $r$ allows emission for only one,resulting in the lowest saturation current. Thus,the graph showing $I_p > I_q > I_r$ is correct.

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