Question

Two metals (Metal A and Metal B) are both exposed to a laser which emits photons that have a greater energy than the work function of both metals. Metal A has a smaller work function than Metal B. Therefore, The de Broglie wavelength of the electrons produced from metal A will be ____________ those from metal B.
The number of electrons produced from metal A will be ____________ the number produced from metal B.

Answer 1

Answer:

shorter than

equal to

Explanation:

Let us go back to the Einstein photoelectric equation;

KE = E - Wo

Where

KE = kinetic energy of the photoelectron

E = energy of the incident photon

Wo = work function of the metal

But KE = 1/2mv^2

Thus the velocity of the emitted photoelectron  is determined by the kinetic energy of the emitted photoelectron.

Since the work function of metal A is smaller than that of metal B, the kinetic energy of photoelectrons emitted from metal A is greater than that of photoelectrons emitted from metal B . Therefore, the velocity of electrons from metal A is greater than those from metal B.

From de Broglie relation;

λ = h/mv

Where;

λ = de Broglie wavelength

h = Plank's constant

m = mass

v = velocity

Metal A producing electrons with greater velocity will lead to a shorter de Broglie wavelength compared to those from metal B.

The number of photoelectrons ejected is determined by the intensity of the photons and not the energy of the incident photons or the work function of the metals. Since the two metals are exposed to the the same laser, equal number of photoelectrons are produced for metals A and B.