- #1

instantresults

- 11

- 1

## Homework Statement

A vacuum photodiode detector utilizes the photoelectric effect to detect light. The photoelectric effect causes electrons to be ejected from a metal plate when photons of light are absorbed by the metal. The energy of a photon is given by the equation

*E*= h

*f*, where h = 6.6 x 10

^{-34}J·s (Planck’s constant), and

*f*is the frequency of the photon. To free an electron, the energy of a photon must be greater than a quantity called the

*work function*of the metal. The ejected electron will have a kinetic energy equal to the photon’s energy minus the work function.

A vacuum photodiode is constructed by sealing two electrodes, a cathode and an anode, in a vacuum tube. The electrodes are separated by a distance,

*L*= 0.01 m, and connected to a battery and a resistor,

*R*= 100 Ω, as shown in Figure 1. The cathode is made of a photoelectric metal and is connected to the negative terminal of the battery. The potential difference between the cathode and anode is approximately equal to the battery voltage,

*V*= 50 V. The electric field at all points between the electrodes is equal to the electrode voltage difference divided by

*L*. The potential energy of an electron immediately after it is released from the cathode is equal to q

*V*, where q = -1.6 x 10

^{-19}C is the charge of an electron. The work function for the vacuum photodiode is 2 x 10

^{-19}J.

Which of the following changes to the circuit will decrease the electric field between the electrodes by the greatest amount?

a) Increasing L by a factor of 2

b) Decreasing L by a factor of 2

c) Increasing R by a factor of 2

d) Decreasing R by a factor of 2

## Homework Equations

V=IR

**E**= V/L

## The Attempt at a Solution

FYI, there are 5 questions attached to this question stem, I'm just stuck on this first part.

So if L = 01m and V = 50V then

**E**

_{initial}= V/L = 50V/.01m = 5000N/C

also, I = 50V/100Ω = 0.5A

a) E when increasing L by factor of 2 = 50V/(.01m*2) = 2500 N/C (decreases by 2500 N/C)

b) E when decreasing L by factor of 2 = 50V/(.01m/2) = 10000 N/C (increases by 5000 N/C)

c) E when increasing R by factor of 2 = (0.5A*200Ω) / .01m = 10000 N/C (increases by 5000 N/C)

d) E when decreasing R by factor of 2 = (0.5A*50Ω) / .01m = 2500 N/C (decreases by 2500 N/C)

The answer is (a) , but I can't figure out why answer (d) also gives a 2500 decrease. I am obviously doing something wrong.

The answer hint states that E= (V-IR)/L but I'm confused how they get that. Thanks for any help