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

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The stopping potential increases.

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = h
u - \Phi$,where $\Phi$ is the work function of the metal.Since $
u = c / \lambda$,reducing the wavelength $\lambda$ increases the frequency $
u$ of the incident radiation.An increase in frequency $
u$ leads to an increase in the maximum kinetic energy $K_{max}$ of the photoelectrons.Since the stopping potential $V_s$ is related to the maximum kinetic energy by $K_{max} = eV_s$,an increase in $K_{max}$ results in an increase in the stopping potential $V_s$.Therefore,the stopping potential increases.

In a photoelectric experiment,if the wavelength of incident radiation is reduced from $6000 \ \mathring{A}$ to $4000 \ \mathring{A}$ while keeping the intensity of radiation constant,then:

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In the experiment of $P.E.E.$ (Photoelectric Effect), the saturation current is $5\,mA$ and the stopping potential is $10\,V$. If the intensity and frequency of the incident light are both doubled, then what will be the new saturation current $(i_s)$ and stopping potential $(V_s)$?

The stopping potential $(V_0)$ versus frequency $(\nu)$ plot of a substance is shown in the figure. The threshold wavelength is:

From a metallic surface,photoelectric emission is observed for frequencies $v_1$ and $v_2$ $(v_1 > v_2)$ of the incident light. The maximum kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1:x$. Hence,the threshold frequency of the metallic surface is

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