Question
The graph shows variation of stopping potential $V_{o}$ with the frequency $v$ of the incident radiation for three photosensitive metals $X_{1},X_{2}$ and $X_{3}$. Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation?
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Solution
Sol. [Key Concept: $KE = h\nu - \varphi$; the metal with the smallest work function (lowest threshold) emits electrons of greatest $KE$.]
$$h\nu = \phi + {ev}_{0}$$
$$\begin{matrix} & v_{0} = \frac{hv}{e} - \frac{\phi}{e} \Rightarrow 0 \Rightarrow \frac{\text{ }hv}{e} = \frac{\phi}{e} \\ & \begin{pmatrix} \phi_{1}:\phi_{2}:\phi_{3} \\ v_{1}:v_{2}:v_{3} \\ 1:1.5:2 \end{pmatrix} \end{matrix}$$
As metal (1) having minimum value of work function
So $x_{1}$ will having maximum kinetic energy
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