In a photoelectric effect experiment,three di | vedclass

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(D) The energy of incident photons is given by $E = \frac{hc}{\lambda}$.For plate $p$: $E_p = \frac{1240}{550} \approx 2.25 \ eV$. Stopping potential

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(D) The energy of incident photons is given by $E = \frac{hc}{\lambda}$.For plate $p$: $E_p = \frac{1240}{550} \approx 2.25 \ eV$. Stopping potential $V_p = E_p - \phi_p = 2.25 - 2.0 = 0.25 \ eV$.For plate $q$: $E_q = \frac{1240}{450} \approx 2.75 \ eV$. Stopping potential $V_q = E_q - \phi_q = 2.75 - 2.5 = 0.25 \ eV$.For plate $r$: $E_r = \frac{1240}{350} \approx 3.54 \ eV$. Stopping potential $V_r = E_r - \phi_r = 3.54 - 3.0 = 0.54 \ eV$.Since the intensities are equal,the saturation current $I_s$ is the same for all three plates.Comparing the stopping potentials: $V_p = V_q = 0.25 \ eV$ and $V_r = 0.54 \ eV$.Thus,plates $p$ and $q$ should have the same stopping potential (intercept on the negative $V$-axis),while plate $r$ should have a larger magnitude of stopping potential (further to the left).Looking at the provided graphs,graph $D$ shows $p$ and $q$ starting at the same point on the negative $V$-axis,and $r$ starting further to the left,which matches our calculation.

List-$I$	List-$II$
$A$. $E = hv$	$I$. de Broglie wavelength
$B$. Diffraction and Interference	$II$. Particle nature of light
$C$. $\lambda = h/p$	$III$. Wave nature of light
$D$. Compton effect	$IV$. Energy of photon

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