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(D) The graph shows the variation of stopping potential $V_s$ with frequency $f$ of incident light.According to Einstein's photoelectric equation: $hf

Vedclass · Accepted Answer

$2.10$

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(D) The graph shows the variation of stopping potential $V_s$ with frequency $f$ of incident light.According to Einstein's photoelectric equation: $hf = \phi + eV_s$,where $\phi$ is the work function.At the threshold frequency $f_0$,the stopping potential $V_s = 0$.From the graph,the intercept on the frequency axis (where $V_s = 0$) is $f_0 = 5 	imes 10^{14} \, Hz$.The work function $\phi$ is given by $\phi = hf_0$.Substituting the values: $\phi = (6.62 	imes 10^{-34} \, J \cdot s) 	imes (5 	imes 10^{14} \, Hz) = 33.1 	imes 10^{-20} \, J$.To convert this energy into electron-volts $(eV)$,divide by the charge of an electron $(e = 1.6 	imes 10^{-19} \, C)$:$\phi = \frac{33.1 	imes 10^{-20}}{1.6 	imes 10^{-19}} \, eV = 2.06875 \, eV \approx 2.07 \, eV$.Given the options,the closest value is $2.10 \, eV$.

The given figure shows a few data points in a photoelectric effect experiment for a certain metal. The minimum energy for the ejection of an electron from its surface is $....... eV$. (Planck's constant $h = 6.62 \times 10^{-34} \, J \cdot s$)

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