The work functions for two different metals A and B,are 3.23 eV and 5.33 eV,respectively.If photons of 220 nm are incident on the two metals,calculate the difference in energy of the fastest photoelectrons from metal A and metal B.
A)0.69 eV
B)0.0 eV
C)2.10 eV
D)1.72 eV
E)2.41 eV

Question

Quizplus · Accepted Answer

The energy of a photon can be calculated using the formula:

E = hc/λ

where h is the Planck's constant, c is the speed of light, and λ is the wavelength of the photon.

For 220 nm, we have:

E = (6.626 x 10^-34 J.s) x (3.0 x 10^8 m/s) / (220 x 10^-9 m)

E = 9.01 x 10^-19 J

We can convert this energy to electron volts (eV) by dividing it by the fundamental charge (e):

E = 9.01 x 10^-19 J / 1.602 x 10^-19 C

E = 5.63 eV

For metal A, the maximum kinetic energy of a photoelectron is:

Kmax = E - Φ

where Φ is the work function of the metal.

Kmax = 5.63 eV - 3.23 eV = 2.40 eV

For metal B, the maximum kinetic energy of a photoelectron is:

Kmax = E - Φ

Kmax = 5.63 eV - 5.33 eV = 0.30 eV

The difference in energy of the fastest photoelectrons from metal A and metal B is:

ΔKmax = Kmax(A) - Kmax(B)

ΔKmax = 2.40 eV - 0.30 eV = 2.10 eV

Therefore, the correct answer is C) 2.10 eV.

The Work Functions for Two Different Metals a and B,are