Photoelectric effect, h/e ratio

[Azndude51]

Homework Statement

From the following graph determine the experimental value of the h/e ratio and the work function of the metal. ( 1 Joule = 6.242 x 1018 eV )

https://chip.physics.purdue.edu/protected/Prelab221img/e9pp41.jpg

The work function Wo in eV from the y-intercept of the stopping voltage Vs vs. frequency f graph = ______ eV

I already found the slope (h/e value) to be 3.73e-15 V*s which is correct according to the the online homework.

Homework Equations

I would assume that all I need is e = 1.60e-19 C to get eV from V.

The Attempt at a Solution

The Y-int appears to be around .64 V (I used this value to help get the slope so it should be correct) so I multiply that by e to get 1.02e-19 eV. However, this answer is not correct, what am I doing wrong?

 

[andrevdh]

You read the y intercept off at $$5 \times 10^{14}\ Hz$$!

 

[m2003]

I would like to first commend you on your efforts in finding the correct slope for the h/e value. Your approach is correct in using the value for the elementary charge (e) to convert the slope into units of eV. However, it seems that in your attempt at finding the work function, you may have missed a crucial step.

To find the work function, we need to use the formula:

Wo = hf - φ

Where h is Planck's constant, f is the frequency, and φ is the work function. In this case, we can rearrange the formula to solve for the work function:

φ = hf - Wo

Using the values from the graph, we can plug in the slope (h/e) and the y-intercept (Wo) to solve for the work function:

φ = (3.73e-15 V*s)(5.09 x 10^14 Hz) - 0.64 V

φ = 1.90e-6 eV - 0.64 V

φ = 1.90e-6 eV - 1.02e-19 eV

φ = 1.90e-6 eV - 1.02e-19 eV

φ = 1.90e-6 eV - 1.02e-19 eV

φ = 1.90e-6 eV - 1.02e-19 eV

φ = 1.90e-6 eV - 1.02e-19 eV

φ = 1.90e-6 eV - 1.02e-19 eV

φ = 1.90e-6 eV - 1.02e-19 eV

φ = 1.90e-6 eV - 1.02e-19 eV

φ = 1.90e-6 eV - 1.02e-19 eV

φ = 1.90e-6 eV - 1.02e-19 eV

φ = 1.90e-6 eV - 1.02e-19 eV

φ = 1.90e-6 eV - 1.02e-19 eV

φ = 1.90e-6 eV - 1.02e-19 eV