The graph shows the variation of stopping pot | vedclass

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(A) According to Einstein's photoelectric equation,$K_{max} = h
u - \phi = \frac{hc}{\lambda} - \phi$.For the same wavelength $\lambda$,the energy of

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$X_1$

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(A) According to Einstein's photoelectric equation,$K_{max} = h
u - \phi = \frac{hc}{\lambda} - \phi$.For the same wavelength $\lambda$,the energy of the incident photon $\frac{hc}{\lambda}$ is constant.The maximum kinetic energy $K_{max}$ is higher for the metal with the smaller work function $\phi$.The work function is given by $\phi = h
u_0$,where $
u_0$ is the threshold frequency (the $x$-intercept of the graph).From the graph,$X_1$ has the lowest threshold frequency $(
u_0 = 1.0 	imes 10^{14} 	ext{ Hz})$,which implies it has the smallest work function.Therefore,for a fixed incident wavelength,metal $X_1$ will emit photoelectrons with the greatest maximum kinetic energy.

The graph shows the variation of stopping potential $V_o$ with the frequency $\nu$ of the incident radiation for three photosensitive metals $X_1, X_2$ and $X_3$. Which metal will emit photoelectrons with greater kinetic energy,for the same wavelength of incident radiation?

Similar Questions

$A$ photon of energy $hv$ is absorbed by a free electron of a metal having work function $\phi < hv$.

The work functions of three metals $A, B$ and $C$ are $W_A, W_B$ and $W_C$ respectively. They are in decreasing order $(W_A > W_B > W_C)$. The correct graph between the maximum kinetic energy $E_k$ of the emitted electron and the frequency $v$ of the incident radiation is:

The slope of the graph between stopping potential and the frequency of incident radiation for the photoelectric effect is . . . . . . .

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