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(B) From the given graph,the photoelectric current becomes zero at a voltage of $V = -4 \ V$. This voltage is known as the stopping potential,denoted

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$4 \ eV$

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(B) From the given graph,the photoelectric current becomes zero at a voltage of $V = -4 \ V$. This voltage is known as the stopping potential,denoted by $V_s$.Therefore,the stopping potential $V_s = -4 \ V$,and its magnitude is $|V_s| = 4 \ V$.The maximum kinetic energy $(K_{max})$ of the emitted photoelectrons is related to the stopping potential by the formula:$K_{max} = e|V_s|$Substituting the value of $|V_s|$,we get:$K_{max} = e 	imes 4 \ V = 4 \ eV$.Thus,the maximum kinetic energy of the emitted photoelectrons is $4 \ eV$.

The figure represents the graph of photoelectric current $I$ versus applied voltage $(V)$. The maximum kinetic energy of the emitted photoelectrons is

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