From the figure describing the photoelectric | vedclass

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Question

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = h
u - \Phi_0

Vedclass · Accepted Answer

Maximum kinetic energy for both the metals depends linearly on the frequency.

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = h
u - \Phi_0$,where $h$ is Planck's constant,$
u$ is the frequency of incident light,and $\Phi_0$ is the work function of the metal.Since the stopping potential $V_s$ is related to the maximum kinetic energy by $eV_s = K_{max}$,we have $V_s = \frac{h}{e}
u - \frac{\Phi_0}{e}$.This equation represents a straight line with a slope of $\frac{h}{e}$.In the given graph,the stopping potential $V_s$ is plotted against frequency $
u$ for two metals,$Na$ and $Al$.Since both lines are straight and parallel (having the same angle $	heta$),it confirms that the maximum kinetic energy for both metals depends linearly on the frequency of the incident radiation.

From the figure describing the photoelectric effect,we may infer correctly that:

Similar Questions

Photoelectrons are ejected from a metal when light of frequency $v$ falls on it. Pick out the wrong statement from the following:

If the frequency of incident light falling on a metallic surface is doubled, what happens to the maximum kinetic energy of the emitted photoelectrons?

Photoelectrons are emitted from two similar metal plates when wavelengths $\lambda_1$ and $\lambda_2$ are incident on them $(\lambda_1 = 1.5\lambda_2)$. If the maximum kinetic energies of the emitted photoelectrons are $E_1$ and $E_2$ respectively, then:

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