How Does Frequency Affect Electron Ejection in the Photoelectric Effect?

[Marghk]

Homework Statement

Having some problem with this question:

The threshold fequency for a material is 5.0*10^14 Hz. Find (a) the work function and (b) the maximum velocity of electrons ejected by radiation of frequency 8.0*10^14 Hz. The mass of an electron is 9.1*10^-31kg.

Homework Equations

(a) Can easily be found with W = hf.
: 6.62*10^-34*5*10^14
= 33.1*10^-20J

My teacher then asked me to calculate the above answer into Electron Volts, which I'm unsure of the answer.

(b) is also giving me some problems. I'm using the equation KE = hf - W, but I don't know where to finish the equation. Below is my attempt.

The Attempt at a Solution

KE(max) = hf - W
= (6.62*10^-34*8*10^14) - (33.1*10^-20)
= 19.86*10^-20J

After this, I get stuck :S I know I need to incorporate the mass of the electron in, but I can't figure it out. If anyone could explain this, it would be great :D

 

[Mindscrape]

So the Joules to eV is pretty simple, so I'm going to omit it, though still feel free to ask for help. 1 electron volt = 1.60217646 × 10-19 joules

The mass of the electron is conveniently placed in units of eV/(m/s)^2, specifically for working in eV. The electron mass, in this form, is given by 0.511 MeV/c^2. Does this help?

 

[m2003]

The Photoelectric Effect is a phenomenon in which electrons are emitted from a material when it is exposed to electromagnetic radiation. In order to understand the problem given, we need to use the equation for the Photoelectric Effect, which is KE = hf - W, where KE is the kinetic energy of the emitted electron, hf is the energy of the incident radiation, and W is the work function of the material.

(a) To find the work function, we can use the equation W = hf, where h is the Planck's constant and f is the threshold frequency of the material. Plugging in the values, we get W = (6.62*10^-34 J*s)(5.0*10^14 Hz) = 3.31*10^-19 J.

Since the work function is usually expressed in electron volts (eV), we can convert the answer by using the conversion factor 1 eV = 1.6*10^-19 J. Therefore, the work function is 3.31*10^-19 J / (1.6*10^-19 J/eV) = 2.07 eV.

(b) To find the maximum velocity of the ejected electrons, we need to use the equation KE = 1/2mv^2, where m is the mass of the electron and v is the velocity. We already know the value of KE from part (a), which is 19.86*10^-20 J. Plugging in the values, we get 19.86*10^-20 J = 1/2(9.1*10^-31 kg)v^2.

Solving for v, we get v = √(2*19.86*10^-20 J / 9.1*10^-31 kg) = 4.5*10^6 m/s.

Therefore, the maximum velocity of the ejected electrons is 4.5*10^6 m/s.