(c) A photo emissive surface has a threshold frequency of 4.02 x 10\(^{14}\)Hz. If the surface is illuminated by light of frequency 5.0 x 10\(^{15}\)Hz, calculate the:
(iii) kinetic energy of the emitted photoelectrons. [ c = 3.0 x 10\(^{8}\) ms\(^{-1}\), h = 6.63 x 10\(^{-34}\) Js]
(a)(i) Einstein's photoelectric equation:
\[ hf = W_0 + \tfrac{1}{2}mv_{\text{max}}^2 \quad\text{(or } hf = hf_0 + E_{k(\max)}\text{)} \]
where \(hf\) is the energy of the incident photon, \(W_0 = hf_0\) the work function of the surface, and \(\tfrac{1}{2}mv_{\max}^2\) the maximum kinetic energy of the emitted photoelectron.
(a)(ii) The equation represents the principle of conservation of energy.
(b) Three applications of photocells: automatic (electric-eye) door openers and burglar/fire alarms; light meters (exposure meters) in cameras; reproduction of sound from the sound track of cinema film (also acceptable: automatic street lighting, television camera tubes).
(c) \(f_0 = 4.02\times10^{14}\,\text{Hz}\), \(f = 5.0\times10^{15}\,\text{Hz}\), \(c = 3.0\times10^{8}\,\text{ms}^{-1}\), \(h = 6.63\times10^{-34}\,\text{Js}\).
(i) Threshold wavelength:
\[ \lambda_0 = \frac{c}{f_0} = \frac{3.0\times10^{8}}{4.02\times10^{14}} = 7.46\times10^{-7}\,\text{m} \]
(ii) Work function:
\[ W_0 = hf_0 = (6.63\times10^{-34})(4.02\times10^{14}) = 2.67\times10^{-19}\,\text{J}\ (\approx 1.67\,\text{eV}) \]
(iii) Kinetic energy of emitted photoelectrons:
\[ E_k = h(f - f_0) = (6.63\times10^{-34})(5.0\times10^{15} - 4.02\times10^{14}) \]
\[ E_k = (6.63\times10^{-34})(4.598\times10^{15}) = 3.05\times10^{-18}\,\text{J} \]