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Chapter 34: Problem 46

The stopping potential in a photoelectric experiment is \(1.8 \mathrm{V}\) when the illuminating radiation has wavelength \(365 \mathrm{nm}\). Determine (a) the work function of the emitting surface and (b) the stopping potential for 280 -nm radiation.

Short Answer

Expert verified
The work function of the emitting surface is approximately \(4.2 \times 10^{-19} \mathrm{J}\) and the stopping potential for 280 nm radiation is about \(2.2 \mathrm{V}\).

Step by step solution

01

Understanding the photoelectric equation

Einstein’s photoelectric equation is \(E = hf = \phi + K_{max}\), where \(E\) is the energy of the incident light, \(hf\) is the energy of the photons (with \(h\) being Planck’s constant and \(f\) the frequency of the radiation), \(\phi\) is the work function (energy necessary to remove an electron) and \(K_{max}\) is the maximum kinetic energy of the emitted electrons. The term \(eV_0 = K_{max}\) represents the retarding potential related energy, with \(e\) the electron charge and \(V_0\) the stopping potential.
02

Calculate the work function

Rewrite the equation and solve for the work function \(\phi\), this gives \(\phi = hf - eV_0\). For light, the frequency can be expressed in terms of wavelength \(\lambda\) as \(f = c/\lambda\), with \(c\) being the speed of light. Substituting this into the equation gives \(\phi = hc/\lambda - eV_0\). Substitute the given values: \(h = 6.626 \times 10^{-34} \mathrm{Js}, e = 1.6 \times 10^{-19} \mathrm{C}, V_0 = 1.8 \mathrm{V}, c = 3 \times 10^8 \mathrm{m/s}\), and \(\lambda = 365 \times 10^{-9}\) m. Doing the calculation yields a value for \(\phi\).
03

Calculate the stopping potential for 280-nm radiation

For step (b), use the calculated work function and the formula \(V_0 = hf/e - \phi/e\). Substitute the value for frequency as done before. Use the value of \(\lambda = 280 \times 10^{-9} m\). After doing the calculation, get the new stopping potential.

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