Photoelectric Question (involving a photocell)

[slinky102]

Homework Statement

Monochromatic light of constant intensity falls on a photocell. The graph shows how the current in the photocell varies with the potential difference applied across it.

The frequency of the incident light is 6.0 x 10^14 Hz. Use the graph to estimate the work function of the metal which forms the cathode of the photocell.

Work function = ___

Add to the axes above the graph obtained when only the intensity of the light is increased. Label this graph A.
Add to the axes above the graph obtained when only the frequency of the light is increased. Label this B.

Homework Equations

E=hf
Work function = hF0
1eV = 1.6 x 10^-19
Plancks constant (h) = 6.63 x 10^-34

The Attempt at a Solution

Have not attempted because I have no idea where to start or go around getting the solution. And could you explain what I would do for the two graph questions. Thanks for any help in advanced. This is for As level physics.

 

[Asim Riaz]

So, you are trying to estimate the work function of the metal that forms the cathode of the photocell. The work function is the minimum amount of energy needed for an electron to escape from the metal surface.To do this, we need to use the equation E=hf, where E is the energy of the incident light, h is Plancks constant and f is the frequency of the incident light. Since we know the frequency of the incident light (6.0 x 10^14 Hz) we can calculate the energy of the incident light:E = hf = 6.63 x 10^-34 x 6.0 x 10^14 = 3.98 x 10^-19 JNow that we know the energy of the incident light, we can use this to estimate the work function. The equation for the work function is Work function = hF0, where F0 is the frequency at which the work function is equal to the energy of the incident light. Since we know the energy of the incident light (3.98 x 10^-19 J), we can solve for F0:Work function = hF03.98 x 10^-19 = 6.63 x 10^-34 F0F0 = 5.99 x 10^14 HzNow we have the frequency at which the work function is equal to the energy of the incident light (5.99 x 10^14 Hz). To find the work function, we just need to multiply this by Plancks constant (6.63 x 10^-34):Work function = hF0 = 6.63 x 10^-34 x 5.99 x 10^14 = 4.09 x 10^-19 J = 2.56 eVSo, the work function of the metal which forms the cathode of the photocell is 2.56 eV.For the two graph questions, Graph A would be the graph obtained when only the intensity of the light is increased, and Graph B would be the graph obtained when only the frequency of the light is increased.