A brief note on Dual Nature of Radiation and Matter . Read it before a test. Try each question at the end before you open its answer.
The photoelectric effect Photocurrent: at a large positive collector voltage it is in proportion to the intensity (the saturation current). Stopping potential V 0 V_0 V 0 : the collector voltage − V 0 -V_0 − V 0 at which the current is zero. K max = e V 0 K_{\max} = eV_0 K m a x = e V 0 . Frequency: V 0 V_0 V 0 rises in a straight line with the frequency, and does not depend on the intensity. Threshold frequency ν 0 \nu_0 ν 0 : below it no electron leaves, whatever the intensity. No delay: emission starts in less than 10 − 9 10^{-9} 1 0 − 9 s. The wave theory of light fails on all three: the energy, the threshold and the delay. Einstein's equation Photon energy: E = h ν = h c / λ E = h\nu = hc/\lambda E = h ν = h c / λ , with h c = 1240 hc = 1240 h c = 1240 eV nm. Equation: K max = h ν − ϕ K_{\max} = h\nu - \phi K m a x = h ν − ϕ , and at the threshold h ν 0 = ϕ h\nu_0 = \phi h ν 0 = ϕ . Graph of V 0 V_0 V 0 against ν \nu ν : a straight line of slope h / e = 4.14 × 10 − 15 h/e = 4.14 \times 10^{-15} h / e = 4.14 × 1 0 − 15 V s, cutting the ν \nu ν axis at ν 0 \nu_0 ν 0 and the V 0 V_0 V 0 axis at − ϕ / e -\phi/e − ϕ / e . Photons each second: N = P / E N = P/E N = P / E . Photocurrent: I = η N e I = \eta Ne I = η N e . Matter waves Photon momentum: p = h / λ p = h/\lambda p = h / λ . A photon has no rest mass. De Broglie wavelength: λ = h / p = h / m v \lambda = h/p = h/mv λ = h / p = h / m v . At a given energy, λ = h / 2 m K \lambda = h/\sqrt{2mK} λ = h / 2 m K . Electrons speeded up by V V V volt: λ = 1.227 / V \lambda = 1.227/\sqrt{V} λ = 1.227/ V nm. Davisson and Germer: electrons diffract from a nickel crystal. Peaks at d sin φ = n λ d\sin\varphi = n\lambda d sin φ = nλ . Waves show only when a gap is about as wide as the wavelength. A worked example Light of wavelength 413 413 413 nm falls on a metal with a work function of 1.9 1.9 1.9 eV. Find the stopping potential.
Photon energy. E = 1240 eV nm 413 nm = 3.0 E = \dfrac{1240\ \text{eV nm}}{413\ \text{nm}} = 3.0 E = 413 nm 1240 eV nm = 3.0 eV. Largest kinetic energy. K max = 3.0 eV − 1.9 eV = 1.1 K_{\max} = 3.0\ \text{eV} - 1.9\ \text{eV} = 1.1 K m a x = 3.0 eV − 1.9 eV = 1.1 eV. Stopping potential. V 0 = 1.1 V_0 = 1.1 V 0 = 1.1 V. Practise with new numbers
Check yourself A photocell is lit by light of the same colour but twice the intensity. What changes: the saturation current, the stopping potential, or both? Stuck? Show a hint Show the answerTwo photons have wavelengths of 450 450 450 nm and 600 600 600 nm. Which has the larger momentum? Stuck? Show a hint Show the answerWhy does a moving cricket ball not show diffraction? Stuck? Show a hint Show the answer