In a photoelectric effect experiment, the stopping voltage \(V_{stop}\) was measured for different frequencies \(f\) of incident light. The relationship between \(V_{stop}\) and \(f\) is given by the equation:
\(V_{stop} = \frac{h}{e} f – \frac{Φ}{e}\)
where \(h\) is Planck’s constant, \(e\) is the elementary charge, and \(Φ\) is the work function of the metal surface.
The experimental data is well represented by the linear equation:
\(V_{stop} = (4.14 \times 10^{-15}) f – 1.20\)
where \(V_{stop}\) is in volts (V) and \(f\) is in hertz (Hz).
Use the elementary charge \(e = 1.60 \times 10^{-19}\) C.
Answer the following:
Question
Determine the threshold frequency \(f_0\) for the metal.
Hint
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The threshold frequency is the frequency at which the stopping voltage is zero.
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Complete Answer
\(f_0 = 2.90 \times 10^{14}\) Hz
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Question
Calculate the work function \(Φ\) of the metal in joules (J).
Hint
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Hint
Relate the y-intercept of the given equation to the term \(-\frac{Φ}{e}\) in the theoretical equation.
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\(Φ = 1.92 \times 10^{-19}\) J
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Question
Using the gradient of the graph, determine the value of Planck’s constant \(h\). Compare your calculated value with the accepted value \(h_{accepted} = 6.63 \times 10^{-34}\) J s.
Hint
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Relate the gradient of the given equation to the term \(\frac{h}{e}\) in the theoretical equation. Calculate h and then compare it to the accepted value.
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\(h = 6.62 \times 10^{-34}\) J s. This value is very close to the accepted value.
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