Tuesday, February 8, 2011

Mastering Physics: The Photoelectric Effect Experiment

Learning Goal: To understand the experiment that led to the discovery of the photoelectric effect.
In 1887, Heinrich Hertz investigated the phenomenon of light striking a metal surface, causing the ejection of electrons from the metal. The classical theory of electromagnetism predicted that the energy of the electrons ejected should have been proportional to the intensity of the light. However, Hertz observed that the energy of the electrons was independent of the intensity of the light. Furthermore, for low enough frequencies, no electrons were ejected, no matter how great the intensity of the light became. The following problem outlines the methods used to investigate this new finding in physics: the photoelectric effect.
Suppose there is a potential difference between the metal that ejects the electrons and the detection device, such that the detector is at a lower potential than the metal. The electrons slow down as they go from higher to lower electric potential; since they must overcome this potential difference to reach the detector, this potential is known as the stopping potential. To reach the detector, the initial kinetic energy of an ejected electron must be greater than or equal to the amount of energy it will lose by moving through the potential difference.
Part A
If there is a potential difference V between the metal and the detector, what is the minimum energy E_min that an electron must have so that it will reach the detector?
Express your answer in terms of V and the magnitude of the charge on the electron, e.
ANSWER:

  E_min  = eV
 For the incident light to cause the ejection of an electron, the light must impart a certain amount of energy to the electron to overcome the forces that constrain it within the metal. The minimum amount of energy required to overcome these forces is called the work function phi. Different metals will have different values for phi. For an electron to reach the detector, the light must impart enough energy for the electron to overcome both the work function and the stopping potential.
Part B
Suppose that the light carries energy E_light. What is the maximum stopping potential V_0 that can be applied while still allowing electrons to reach the detector?
 Express your answer in terms e, E_light, and phi.
ANSWER:

  V_0  = \frac{\left(E_{light}-{\phi}\right)}{e}


Part C
Classical electromagnetism predicted that V_0 should have increased as the intensity of the incident light increased. On the contrary, it was found that V_0 increased as the frequency f of the light increased. The voltage V_0 was found to obey the following linear relationship:
V_{0} = mf - b,
where m and b are numerical constants (representing the slope and the intercept, respectively). By comparing this equation to your answer from Part B, find an expression for the intercept b. (Notice that mf in this equation changes with different light but b is a constant of the metal.)
Express your answer in terms of phi and e.
ANSWER:

  b  = \frac{{\phi}}{e}



Part D
In a 1905 paper that later won him a Nobel Prize, Albert Einstein postulated that the energy of light was proportional to its frequency. The constant of proportionality turned out to be Planck's constant h: E_{\rm light} = hf. Using your previous results, and the equation given in Part C, find an expression for h in terms of experimentally determinable quantities.
Express your answer in terms of the slope m and e.
ANSWER:

  h  = me


Part E
Suppose that two sets of values were recorded in this experiment:
Stopping potential V_0
(\rm V)
Frequency f
(\rm Hz)
0.551 6 \times 10^{14}
0.965 7 \times 10^{14}
Using these data, extrapolate a numerical value for Planck's constant h.
Express your answer in joule-seconds to three significant figures.
ANSWER:

  h  = 6.630×10−34
  \rm J \cdot s
Part F
Using the data given, find a numerical value for the work function phi of the metal.
Express your answer in joules to two significant figures.
ANSWER:

  phi  = 3.10×10−19
    \rm J

4 comments:

  1. You have made my studying infinity easier.

    ReplyDelete
  2. Thank you, stil works (27 April 2016)

    ReplyDelete
  3. Thanks your explanation made it much easier to undestand

    ReplyDelete