# Homework Help: Einstein's photoelectric equation

1. Jul 18, 2007

### t_n_p

1. The problem statement, all variables and given/known data

1. How can Planck's constant and the work function be determined from the following graph?

http://img507.imageshack.us/img507/9076/untitledgo5.jpg [Broken]

2. Calculate the frequencies corresponding to the cut-off wavelengths for filters 15 and 29 in table 1. Take speed of light as 3*10^8 m/s.

http://img515.imageshack.us/img515/2240/untitled2jh5.jpg [Broken]

3. The attempt at a solution

I know the following equations, but can't seem to peice it all together

KE = hf - W

KE = eVs (where Vs is called the stopping potential, the potential for which the photocurrent reaches a value equal to zero)

and Vs = (h/e)f - (W/e)

Last edited by a moderator: May 3, 2017
2. Jul 18, 2007

### rahuldandekar

A hint: the equation of a line which intersects the x axis at A and has slope m is

y = m*(x-A)

(Think about why... the case y=0 )

3. Jul 18, 2007

### andrevdh

1. You can see from your final equation that the gradient of the graph will be h/e and the y-intercept of the graph (which is not shown in the picture) will be (- W/e). Your final equation is a linear function of the frequency of the incident light.

2. The filters transmits colours with longer wavelengths up to the minimum wavelengths stated. Just calculate the frequencies (cut-off ...) for these minimum wavelengths from

$$c = \lambda f$$

This means that these filters will transmit light of lower frequencies up to these calculated ones. Since c is constant in the equation above one need to choose a larger f with a smaller $$\lambda$$ to obtain the same product. That is a short wavelength causes the peaks to pass a point quicker (larger frequency) than a longer wavelength when they both move at the same speed.

4. Jul 18, 2007

### t_n_p

In regards to q1, gradient will be h/e. Therefore h = gradient*e and work function = y-intercept*-e?

Sorry, I'm slightly confused.

5. Jul 19, 2007

### andrevdh

Correct. The first graph plots the stopping potential for the emitted electrons as a function of the frequency of the incident photons on the material. It therefore displays the relationship between Vs and f. That is exactly what your final equation gives - the relationship between Vs and f. Both h/e and W/e are constants in this equation. So this is a linear relationship between Vs and f.

6. Jul 19, 2007

### t_n_p

Gotcha, now in regards to q2.
I get the explanation behind c = lambda*f (i.e. as wavelenth gets smaller, frequency gets larger and vice versa), but how do I actually find the numerical values of the frequencies?

7. Jul 19, 2007

### andrevdh

The minimum wavelengths are given in the table so you just calculate the maximum frequencies with

$$c = \lambda f$$

8. Jul 19, 2007

### GoldPheonix

You forgot some equations.

$$c = \lambda f$$

$$E = h f$$

Where h -> Plank's Cosnt, f -> frequency, and c -> speed of light, E -> energy, lambda -> wavelength