How Does the Photoelectric Effect Determine Planck's Constant?

[sofreshjay]

Homework Statement

In an experiment to determine the value of Planck's constant a negative potential was applied to the anode of a photoelectric cell and the minimum potential required to reduce the photocurrent to zero was mesure for incident light of various frequencies. For the frequencies of 8.5 x 10^14 Hz and 2.35 x 10^15 Hz the minimum potentials required were found to be 0.4V and 6.4V, respectively. Calculate Planck"s constant h.

Homework Equations

E=hf=hc/wavelength
E=mc^2

3. Attempt

I do not know how to start this. IN NEED OF DIRE HELP.
Thank you to anyone who helps.

 

[rock.freak667]

Even though I shouldn't reply since you didn't attempt anything, I will.

but you know that $E=\phi +E_{k(max)}$ which is the same as $hf=hf_0+\frac{1}{2}mv_{max}^2$

When the photocurrent reaches zero, it means that the photoelectrons with max ke. just fail to reach the anode. So that the work done in stopping the electrons with max ke from reaching the anode=Loss in ke of the photoelectrons with max ke.

Can you express the last line in terms of an equation?

 

[Moetasim]

Dear student,

The photoelectric effect is a phenomenon in which electrons are emitted from a material when it is exposed to light of a certain frequency. This effect was first discovered by Albert Einstein and has since been used in various experiments to study the properties of light and matter.

In this experiment, you have measured the minimum potential required to reduce the photocurrent to zero for two different frequencies of incident light. The equations you will need to use to calculate Planck's constant are E=hf and E=mc^2.

First, let's rearrange the equation E=hf to solve for Planck's constant, h. We can write it as h=E/f. We know that the energy, E, is equal to the product of Planck's constant and the frequency, so we can rewrite this as h=E/f=hf/f. This simplifies to h=E/f, which is the equation we will use to calculate h.

Now, we can substitute the values given in the problem into this equation. For the first frequency, 8.5 x 10^14 Hz, the minimum potential required was 0.4V. So, we have h=(0.4V)/(8.5 x 10^14 Hz). We can simplify this by converting volts to joules (since energy is typically measured in joules) and Hz to s^-1 (since frequency is typically measured in cycles per second). This gives us h=(0.4J)/(8.5 x 10^14 s^-1). We can then use the equation E=mc^2 to convert this to the SI unit for energy, which is joules. This gives us h=(0.4kgm^2s^-2)/(8.5 x 10^14 s^-1)=4.7 x 10^-15 kgm^2s^-2.

We can repeat this process for the second frequency, 2.35 x 10^15 Hz, and the minimum potential of 6.4V. This gives us h=(6.4J)/(2.35 x 10^15 s^-1)=2.7 x 10^-15 kgm^2s^-2.

Now, to calculate the average value of Planck's constant, we can take the average of these two values: (4.7 x 10^-15 kgm^2s^-2 + 2.7 x 10^-15