Question

In an experiment electrons are released with a kinetic energy of 2.8 eV. The amazing physicists doing the experiments are increasing the wavelength by 50%. Do you remember what happens? Are the photoelectrons faster or slower? The energy now is 0.44 eV.

A) Based on that information, what is the material of the cathode? (calculate the work function and compare to table 28.1 in our textbook).

B) What is the initial wavelength of the light?

C) When you go a few pages back in the textbook you can find out the temperature of a physical object radiating at that wavelength

Answer 1

Photoelectric effect can be represented by the equation
KE = hc/ - w
KE is the kinetic energy of the electron, is the wavelength of the incident photons and w is the work function, which is the minimum energy needed to remove an electron form the cathode, h is Plank's constant and c is the velocity of the light
Case 1
KE = 2.8 eV and wavelength is ,
so, 2.8 eV = hc/ - w ...(1)
Case 2
KE = 0.44 eV and wavelength is 1.5 (wavelength increased by 50%)
So, 0.44 eV = hc/1.5 - w (2)
Photoelectrons will be slower since its kinetic energy decreased

If we substitute the values of h and c, we will get
hc = 1.98 x 10^-25 Jm
?Since the unit of KE is in eVs, we can convert J.m into eV.nm
hc = 1240 eV nm, [ 1 eV = 1.6 x 10^-19 J and 1 nm = 10^-9 nm.

A)
Multiply eqn (2) with 1.5 and subtracting it from (1), we will get
(1) - 1.5 x (2) gives
2.8 - 1.5 x 0.44 = 0.5 w
w = 4.28 eV
Gold has a work function in this range (4.26