The figure shows the plot of stopping potenti | vedclass

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(A) From Einstein's photoelectric equation,$eV_0 = \frac{hc}{\lambda} - \phi$,which can be rewritten as $V_0 = \frac{hc}{e}(\frac{1}{\lambda}) - \frac

Vedclass · Accepted Answer

$\phi_1 : \phi_2 : \phi_3 = 1 : 2 : 4$

Vedclass · Answer

(A) From Einstein's photoelectric equation,$eV_0 = \frac{hc}{\lambda} - \phi$,which can be rewritten as $V_0 = \frac{hc}{e}(\frac{1}{\lambda}) - \frac{\phi}{e}$.Comparing this with the equation of a straight line $y = mx + c$,the slope is $m = \frac{hc}{e}$,which is constant for all metals,and the x-intercept is $\frac{1}{\lambda_0} = \frac{\phi}{hc}$.From the graph,the threshold values for $(1/\lambda)$ are $(1/\lambda)_1 = 0.001$,$(1/\lambda)_2 = 0.002$,and $(1/\lambda)_3 = 0.004$.Since $\phi = hc(1/\lambda_0)$,we have $\phi_1 : \phi_2 : \phi_3 = (1/\lambda_0)_1 : (1/\lambda_0)_2 : (1/\lambda_0)_3 = 0.001 : 0.002 : 0.004 = 1 : 2 : 4$.Thus,option $A$ is correct.

The figure shows the plot of stopping potential $(V_0)$ versus $(1/\lambda)$ for three different metals. If $\phi$ is the work function,then which of the following is correct?

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