Photons with an energy of 7.52 eV strike a material that has a work function of 4.22 eV.The maximum kinetic energy of the electron emitted from this material is
A)7.5 eV
B)12 eV
C)3.3 eV
D)0.98 eV
E)No electrons are ejected by these photons.

Question

Quizplus · Accepted Answer

The maximum kinetic energy (KE) of the emitted electron can be found using the equation:

KE = hν - Φ

where h is Planck's constant, ν is the frequency of the photon (related to its energy E by E = hν), and Φ is the work function of the material.

In this case, we have:

E = 7.52 eV
Φ = 4.22 eV

Converting to frequency units (Hz) and substituting into the equation, we get:

KE = (6.626 x 10^-34 J s)(7.52 eV/1.602 x 10^-19 J/eV) - (4.22 eV)

Simplifying and converting back to electronvolt (eV) units, we get:

KE = 3.3 eV

Therefore, the maximum kinetic energy of the emitted electron is 3.3 eV, which corresponds to Choice C.

Photons with an Energy of 7