Photoelectric Effects and deBroglie Equation: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Photoelectric Effects and deBroglie Equation. Understanding the underlying formula is key to arriving at the correct answer below:

The energy of the incident photon is $E = \frac{hc}{\lambda} = \frac{12400}{4100} eV \approx 3.02 eV$ . For photoelectric emission to occur, the incident energy must be greater than the work function ( $E > W$ ). Since $3.02 eV$ is greater than the work functions of A ( $1.92 eV$ ) and B ( $2.0 eV$ ) but less than C ( $5 eV$ ), only A and B will emit photoelectrons.