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Date November 2011 Marks available 7 Reference code 11N.3.SL.TZ0.4
Level Standard level Paper Paper 3 Time zone Time zone 0
Command term Define, Sketch, State, and Outline Question number 4 Adapted from N/A

Question

This question is about the photoelectric effect.

The diagram shows the set up of an experiment designed to verify the Einstein model of the photoelectric effect.

The tungsten electrode is positive.

Explain how the maximum kinetic energy of electrons ejected from the positive electrode is determined.

[3]
a.

Light of frequency f is shone onto the tungsten electrode in (a). The potential Vs for which the photoelectric current is zero is recorded for different values of f.

(i) Using the axes below, sketch a graph of how you might expect Vs to vary with f.

(ii) State the Einstein photoelectric equation in a form that relates Vs and f. Define, other than the electron charge, any other symbols that you might use.

(iii) Outline how a graph of Vs against f can be used to find the Planck constant and work function of tungsten.

[7]
b.

The work function of tungsten is 4.5eV. Show that the de Broglie wavelength of an electron that has this energy is about 0.6nm.

[3]
c.

Markscheme

the potential difference is varied (using the potential divider);
until the current registered by the ammeter is zero;
the maximum kinetic energy of the (ejected) electrons is this potential times the electron charge;

a.

(i)

straight line;
with non-zero intercept on f axis;

(ii) Vse=hfhf0 or Vse=hfφ;
f0→the frequency below which no electron emission takes place;
h→the Planck constant;
φ→the minimum energy required to eject an electron from tungsten;
Award [2 max] if the equation is not given.

(iii) Planck constant: slope/gradient of graph×e;
work function: extrapolation to intercept on Vs axis and φ=Vs-intercept×e / when Vs=0, φ=hf so intercept gives f when Vs=0 and φ=h (f-intercept);

b.

use of \(p = \frac{h}{\lambda }\) and \({E_k} = \frac{{{p^2}}}{{2m}}\);
\(\lambda  = \frac{h}{{\sqrt {2{E_{\rm{k}}}m} }}\);
\( = \frac{{6.6 \times {{10}^{ - 34}}}}{{2 \times 4.5 \times 1.6 \times {{10}^{ - 19}} \times 9.1 \times {{10}^{ - 31}}}} = 5.765 \times {10^{ - 10}}{\rm{m}}\);
≈0.6nm

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Additional higher level (AHL) » Topic 12: Quantum and nuclear physics » 12.1 – The interaction of matter with radiation
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