Question

Figure represents the graph of photo-current $I$ versus applied voltage $\left(V\right)$. The maximum energy of the emitted photoelectron is-

• A. $2eV$
• B. $4eV$
• C. $0eV$
• D. $3eV$

Answer 1

The correct option is B $4eV$

We see that for voltage $-4volt$ the current is $zero$ so $magnitude$ of the cutoff voltage will be $V_0=4volt$

So maximum kinetic energy of emitted photoelectron will be $e{V}_0=4eV$

Option B is correct.