Question

A 3.0 eV light is shined on Gold ( Փ = 5.1 eV), silver (Փ =4.26 eV), Cesium (Փ = 2.1 eV), and Platinum (Փ = 6.35). If Փ is the work function of the metal, what metal would have electrons ejected?

Possible Answers:
a) Au
b) Pt
c) Cs
d) Au, Ag, and Pt
e) Ag

Answer 1

The metal that would have electrons ejected when a 3.0 eV light is shined on it is Cesium.

Step-by-step explanation:

The maximum kinetic energy of ejected photoelectrons can be calculated using the equation:

K.E. = hν - Φ

Where K.E. is the maximum kinetic energy, h is Planck's constant (6.626 x 10-34 J·s), ν is the frequency of the incident light, and Φ is the work function of the metal.

To determine which metal would have electrons ejected, we compare the energy of the incident light (given in electron volts, eV) to the work function of each metal. If the energy of the light exceeds the work function, electrons will be ejected.

For the given incident light energy of 3.0 eV:

1. Gold has a work function of 5.1 eV, so electrons will not be ejected.
2. Silver has a work function of 4.26 eV, so electrons will not be ejected.
3. Cesium has a work function of 2.1 eV, so electrons will be ejected.
4. Platinum has a work function of 6.35 eV, so electrons will not be ejected.

Therefore, the only metal that would have electrons ejected is Cesium.