10 exam-style questions on the photoelectric effect, energy levels, photons and wave–particle duality. Tap Show answer for full working.
Calculate the energy of a photon of wavelength \(450\,\mathrm{nm}\). (\(h = 6.63\times10^{-34}\,\mathrm{J\,s}\), \(c = 3.00\times10^{8}\,\mathrm{m\,s}^{-1}\).)
\(E = hc/\lambda = (6.63\times10^{-34}\times3.00\times10^{8})/450\times10^{-9} = 4.42\times10^{-19}\,\mathrm{J}\).
The work function of a metal is \(2.3\,\mathrm{eV}\). Calculate the threshold frequency for the photoelectric effect. (\(h = 4.14\times10^{-15}\,\mathrm{eV\,s}\).)
\(f_{0} = \phi/h = 2.3/(4.14\times10^{-15}) = 5.56\times10^{14}\,\mathrm{Hz}\).
State Einstein’s photoelectric equation and explain the meaning of each term.
\(hf = \phi + \frac12 mv_{\max}^{2}\).
\(hf\): photon energy; \(\phi\): work function (minimum energy to free an electron); \(\frac12 mv_{\max}^{2}\): maximum kinetic energy of emitted photoelectrons.
Light of frequency \(8.0\times10^{14}\,\mathrm{Hz}\) is incident on a metal surface with work function \(1.8\,\mathrm{eV}\). Calculate the maximum kinetic energy of the photoelectrons in eV and the stopping potential. (\(h = 4.14\times10^{-15}\,\mathrm{eV\,s}\).)
\(hf = 4.14\times10^{-15}\times8.0\times10^{14} = 3.31\,\mathrm{eV}\).
\(KE_{\max} = 3.31 - 1.8 = 1.51\,\mathrm{eV}\).
Stopping potential \(V_{s} = KE_{\max}/e = 1.51\,\mathrm{V}\).
An electron in a hydrogen atom makes a transition from the \(n=3\) level to the \(n=2\) level. The energy of the \(n=2\) level is \(-3.40\,\mathrm{eV}\) and of the \(n=3\) level is \(-1.51\,\mathrm{eV}\). Calculate the wavelength of the emitted photon.
\(\Delta E = -1.51 - (-3.40) = 1.89\,\mathrm{eV} = 1.89\times1.6\times10^{-19} = 3.02\times10^{-19}\,\mathrm{J}\).
\(\lambda = hc/\Delta E = (6.63\times10^{-34}\times3\times10^{8})/3.02\times10^{-19} = 6.58\times10^{-7}\,\mathrm{m}\) (\(658\,\mathrm{nm}\)).
Calculate the de Broglie wavelength of an electron accelerated through a potential difference of \(150\,\mathrm{V}\). (Electron mass \(= 9.11\times10^{-31}\,\mathrm{kg}\), \(e = 1.6\times10^{-19}\,\mathrm{C}\), \(h = 6.63\times10^{-34}\,\mathrm{J\,s}\).)
\(\frac12 mv^{2} = eV \Rightarrow p = \sqrt{2meV} = \sqrt{2\times9.11\times10^{-31}\times1.6\times10^{-19}\times150} = 6.61\times10^{-24}\,\mathrm{kg\,m\,s}^{-1}\).
\(\lambda = h/p = 6.63\times10^{-34}/6.61\times10^{-24} = 1.00\times10^{-10}\,\mathrm{m}\).
Explain why the photoelectric effect provides evidence for the particle nature of light, while electron diffraction provides evidence for the wave nature of electrons.
Photoelectric effect: electron emission depends on frequency (threshold), not intensity; KE of electrons independent of intensity but depends on frequency — explained by photons transferring energy quanta \(hf\).
Electron diffraction: electrons produce interference/diffraction patterns, which are wave phenomena; hence electrons have an associated wavelength (de Broglie).
In a photoelectric experiment the stopping potential is \(0.85\,\mathrm{V}\) when light of wavelength \(480\,\mathrm{nm}\) is used. Calculate the work function of the metal in eV and the threshold wavelength. (\(h = 4.14\times10^{-15}\,\mathrm{eV\,s}\), \(c = 3.00\times10^{8}\,\mathrm{m\,s}^{-1}\).)
\(hf = hc/\lambda = (4.14\times10^{-15}\times3\times10^{8})/480\times10^{-9} = 2.59\,\mathrm{eV}\).
\(\phi = hf - eV_{s} = 2.59 - 0.85 = 1.74\,\mathrm{eV}\).
\(\lambda_{0} = hc/\phi = (4.14\times10^{-15}\times3\times10^{8})/1.74 = 7.14\times10^{-7}\,\mathrm{m}\) (\(714\,\mathrm{nm}\)).
The ground state energy of a hydrogen atom is \(-13.6\,\mathrm{eV}\). Calculate the ionisation energy and the energy of a photon emitted when an electron falls from the \(n=4\) level to the ground state. The energy of level \(n\) is \(E_{n} = -13.6/n^{2}\,\mathrm{eV}\).
Ionisation energy \(= 13.6\,\mathrm{eV}\) (energy to go from \(n=1\) to \(n=\infty\)).
\(E_{4} = -13.6/16 = -0.85\,\mathrm{eV}\).
Photon energy \(= E_{4} - E_{1} = -0.85 - (-13.6) = 12.75\,\mathrm{eV}\).
An electron microscope accelerates electrons through \(20\,\mathrm{kV}\). Calculate the de Broglie wavelength of the electrons and explain why this allows much higher resolution than an optical microscope using visible light (\(\lambda\sim500\,\mathrm{nm}\)). (Use \(m_{e}=9.11\times10^{-31}\,\mathrm{kg}\), \(e=1.6\times10^{-19}\,\mathrm{C}\), \(h=6.63\times10^{-34}\,\mathrm{J\,s}\).)
\(p = \sqrt{2meV} = \sqrt{2\times9.11\times10^{-31}\times1.6\times10^{-19}\times2\times10^{4}} = 7.63\times10^{-23}\,\mathrm{kg\,m\,s}^{-1}\).
\(\lambda = h/p = 8.69\times10^{-12}\,\mathrm{m}\).
The electron wavelength is \(\sim10^{4}\) times smaller than visible light, so the diffraction limit is correspondingly smaller, allowing much finer detail to be resolved.