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PHYSICS, M7 2019 VCE 16

Students are studying the photoelectric effect using the apparatus shown in Figure 1.
 

     

Figure 2 shows the results the students obtained for the maximum kinetic energy \((E_{\text{k max }})\) of the emitted photoelectrons versus the frequency of the incoming light.
 

  1. Using only data from the graph, determine the values the students would have obtained for
    1. Planck's constant, \(h\). Include a unit in your answer.  (2 marks)

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    1. the maximum wavelength of light that would cause the emission of photoelectrons.   (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

    1. the work function of the metal of the photocell.   (1 mark)

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  1. The work function for the original metal used in the photocell is \(\phi\).
    On Figure 3, draw the line that would be obtained if a different metal, with a work function of \(\dfrac{1}{2} \phi\), were used in the photocell. The original graph is shown as a dashed line.   (2 marks)
     

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Show Answers Only

a.i.  \(5.4 \times 10^{-15}\ \text{eV s}\)

   ii.  \(811\ \text{nm}\)

  iii.  \(1.9\ \text{eV}\).

b.  

Show Worked Solution

a.i.  Planck’s constant \((h)\):

  • Equal to the gradient of the line when \(E_{\text{k max}}\) is graphed against frequency.

\(\therefore h=\dfrac{\text{rise}}{\text{run}}=\dfrac{1.25-0}{6 \times 10^{14}-3.7\times 10^{-14}}=5.4 \times 10^{-15}\ \text{eV s}\)
 

a.ii.  Max wavelength = minimum frequency of emitted photoelectron.

\(\lambda=\dfrac{c}{f}=\dfrac{3 \times 10^8}{3.7 \times 10^{14}}=811\ \text{nm}\)
 

♦ Mean mark (a)(ii) 44%.

a.iii.  

   

  • The work function is the y-intercept of the graph, so by extending the graph as shown above, the work function is \(1.9\ \text{eV}\).
     

b.   Constructing the new graph:

  • The new \(y\)-intercept for the graph will be \(-0.95\ \text{eV}\)
  • The gradient of the graph will remain the same (Planck’s constant)
     

Filed Under: Light: Quantum Model Tagged With: Band 4, Band 5, smc-3698-10-Photoelectric Effect, smc-3698-40-Find freq/lambda, smc-3698-70-Planck

PHYSICS, M7 2014 HSC 26

  1. Calculate the energy of a photon of wavelength 415 nm.  (2 marks)
  2. An experiment was conducted using a photoelectric cell as shown in the diagram.

The graph plots the maximum kinetic energy of the emitted photoelectrons against radiation frequency for the aluminium surface.

The experiment is planned to be repeated using a voltage of 0.0 V.

Draw a line on the graph to show the predicted results of the planned experiment, and determine the radiation frequency which would produce photoelectrons with a maximum kinetic energy of 1.2 eV using a voltage of 0.0 V.  (3 marks)

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`text(See Worked Solutions)`

Show Worked Solution
a.   `c` `= f lambda`
    `= hf`
    `= (hc)/lambda`
    `= (6.626 xx 10^(-34) Js xx 3 xx 10^8 ms^(-1))/(415 xx 10^(-9)m)`
    `= 4.79 xx 10^(-19) J`

 

b.  

`text(To find intercept)`

`4.1 text(V)` `= 4.1 xx 1.602 xx 10^(-19)\ text(J)` `text(of energy required to be)`
`text(supplied by the photon.)`
  `= 6.56 xx 10^(-19)\ text(J)`  
`hf` `= 6.56 xx 10^(-19)\ text(J)`  
`f` `= (6.56 xx 10^(-19))/(6.626 xx 10^(-34))`  
  `= 9.9 xx 10^14`  

`text(Gradient = same as A1)`

`text{From graph maximum KE(eV)}` `= 1.2\ text(eV)`
`text(Frequency)` `= 12.8\ text(Hz)`
  `= 12.8 xx 10^14\ text(Hz)`

Filed Under: Uncategorized Tagged With: Band 4, Band 5, smc-3698-40-Find freq/lambda, smc-3698-60-Experiments

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