In an experiment on the photoelectric effect, | vedclass

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(A) According to Einstein's photoelectric equation: $eV_0 = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.Rearranging this,we get: $V_0

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Work function of metal $B$ is greater than that of metal $A$.

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(A) According to Einstein's photoelectric equation: $eV_0 = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.Rearranging this,we get: $V_0 = \frac{hc}{e} \left(\frac{1}{\lambda}\right) - \frac{\phi}{e}$.This is the equation of a straight line $y = mx + c$,where the slope $m = \frac{hc}{e}$ is constant for both metals,and the x-intercept is $\frac{1}{\lambda_0} = \frac{\phi}{hc}$.From the graph,the x-intercept for metal $A$ is smaller than the x-intercept for metal $B$. Since the x-intercept is proportional to the work function $\phi$ (i.e.,$\phi = hc \cdot (1/\lambda_0)$),a smaller x-intercept implies a smaller work function.Therefore,$\phi_A < \phi_B$,which means the work function of metal $B$ is greater than that of metal $A$.

In an experiment on the photoelectric effect,a student plots the stopping potential $V_0$ against the reciprocal of the wavelength $1/\lambda$ of the incident light for two different metals $A$ and $B$. These are shown in the figure. Looking at the graphs,you can most appropriately say that:

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