Chapter 11: Dual Nature of Radiation and Matter

Each photon in a monochromatic beam of frequency $$\nu$$ carries an energy $$E = h\nu$$, where $$h = 6.63 \times 10^{-34}\,\mathrm{J\,s}$$ is Planck's constant.

(a) Energy of a photon.

$$E = h\nu = (6.63 \times 10^{-34}\,\mathrm{J\,s}) \times (6.0 \times 10^{14}\,\mathrm{Hz})$$

$$E = 3.98 \times 10^{-19}\,\mathrm{J}$$

Converting to electron volts by dividing by $$1.6 \times 10^{-19}\,\mathrm{J/eV}$$:

$$E = \frac{3.98 \times 10^{-19}}{1.6 \times 10^{-19}}\,\mathrm{eV} = 2.49\,\mathrm{eV}$$

(b) Number of photons emitted per second.

If the source emits $$N$$ photons per second, each carrying energy $$E$$, then the power (energy radiated per second) is $$P = NE$$. Therefore

$$N = \frac{P}{E} = \frac{2.0 \times 10^{-3}\,\mathrm{W}}{3.98 \times 10^{-19}\,\mathrm{J}}$$

$$N \approx 5.0 \times 10^{15}\ \text{photons per second}$$

For a photosensitive metal of work function $$\phi_0$$, the threshold frequency $$\nu_0$$ is the minimum frequency of incident light that just frees an electron, defined by $$h\nu_0 = \phi_0$$.

(a) Threshold frequency.

First express the work function in joules:

$$\phi_0 = 2.14\,\mathrm{eV} = 2.14 \times 1.6 \times 10^{-19}\,\mathrm{J} = 3.42 \times 10^{-19}\,\mathrm{J}$$

$$\nu_0 = \frac{\phi_0}{h} = \frac{3.42 \times 10^{-19}\,\mathrm{J}}{6.63 \times 10^{-34}\,\mathrm{J\,s}}$$

$$\nu_0 = 5.16 \times 10^{14}\,\mathrm{Hz}$$

(b) Wavelength of the incident light.

A stopping potential $$V_0 = 0.60\,\mathrm{V}$$ means the maximum kinetic energy of the photoelectrons is $$K_{\max} = eV_0 = 0.60\,\mathrm{eV}$$. Einstein's photoelectric equation gives

$$h\nu = \phi_0 + eV_0$$

so the energy of an incident photon is

$$h\nu = (2.14 + 0.60)\,\mathrm{eV} = 2.74\,\mathrm{eV} = 2.74 \times 1.6 \times 10^{-19}\,\mathrm{J} = 4.38 \times 10^{-19}\,\mathrm{J}$$

The incident frequency is

$$\nu = \frac{4.38 \times 10^{-19}}{6.63 \times 10^{-34}} = 6.61 \times 10^{14}\,\mathrm{Hz}$$

and the corresponding wavelength is

$$\lambda = \frac{c}{\nu} = \frac{3.0 \times 10^{8}\,\mathrm{m/s}}{6.61 \times 10^{14}\,\mathrm{Hz}} = 4.54 \times 10^{-7}\,\mathrm{m} \approx 454\,\mathrm{nm}$$

The de Broglie wavelength of a particle of mass $$m$$ moving with speed $$v$$ (momentum $$p = mv$$) is

$$\lambda = \frac{h}{p} = \frac{h}{mv}$$

(a) Electron. With $$m_e = 9.11 \times 10^{-31}\,\mathrm{kg}$$ and $$v = 5.4 \times 10^{6}\,\mathrm{m/s}$$, the momentum is

$$p = m_e v = (9.11 \times 10^{-31}\,\mathrm{kg}) \times (5.4 \times 10^{6}\,\mathrm{m/s}) = 4.92 \times 10^{-24}\,\mathrm{kg\,m/s}$$

$$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}\,\mathrm{J\,s}}{4.92 \times 10^{-24}\,\mathrm{kg\,m/s}} = 1.35 \times 10^{-10}\,\mathrm{m} = 0.135\,\mathrm{nm}$$

This is comparable to atomic dimensions, which is why electrons of this speed display pronounced wave (diffraction) effects.

(b) Ball. With $$m = 150\,\mathrm{g} = 0.150\,\mathrm{kg}$$ and $$v = 30.0\,\mathrm{m/s}$$,

$$p = mv = 0.150 \times 30.0 = 4.50\,\mathrm{kg\,m/s}$$

$$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{4.50} = 1.47 \times 10^{-34}\,\mathrm{m}$$

This wavelength is so vanishingly small (far smaller than any atomic or nuclear size) that the wave nature of an everyday object such as a ball can never be detected.

The cut-off (stopping) voltage $$V_0$$ is the potential needed to stop the most energetic photoelectrons. The work done by this retarding potential in stopping an electron equals the electron's maximum kinetic energy:

$$K_{\max} = eV_0$$

With $$V_0 = 1.5\,\mathrm{V}$$ and $$e = 1.6 \times 10^{-19}\,\mathrm{C}$$:

$$K_{\max} = (1.6 \times 10^{-19}\,\mathrm{C}) \times (1.5\,\mathrm{V})$$

$$K_{\max} = 2.4 \times 10^{-19}\,\mathrm{J}$$

Equivalently, expressed in electron volts, $$K_{\max} = eV_0 = 1.5\,\mathrm{eV}$$.

According to Einstein's photoelectric equation, the cut-off (stopping) potential $$V_0$$ is related to the frequency $$\nu$$ of the incident light by

$$eV_0 = h\nu - \phi_0 \quad\Rightarrow\quad V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}$$

This is the equation of a straight line of $$V_0$$ against $$\nu$$, whose slope is $$\dfrac{h}{e}$$. Therefore

$$\text{slope} = \frac{h}{e} \quad\Rightarrow\quad h = e \times \text{slope}$$

$$h = (1.6 \times 10^{-19}\,\mathrm{C}) \times (4.12 \times 10^{-15}\,\mathrm{V\,s})$$

$$h = 6.59 \times 10^{-34}\,\mathrm{J\,s}$$

This is in excellent agreement with the accepted value $$h = 6.63 \times 10^{-34}\,\mathrm{J\,s}$$.

Einstein's photoelectric equation can be written in terms of the threshold frequency $$\nu_0$$, since the work function is $$\phi_0 = h\nu_0$$:

$$eV_0 = h\nu - h\nu_0 = h(\nu - \nu_0)$$

Hence the cut-off potential is

$$V_0 = \frac{h(\nu - \nu_0)}{e}$$

Substituting $$\nu = 8.2 \times 10^{14}\,\mathrm{Hz}$$ and $$\nu_0 = 3.3 \times 10^{14}\,\mathrm{Hz}$$:

$$\nu - \nu_0 = (8.2 - 3.3) \times 10^{14}\,\mathrm{Hz} = 4.9 \times 10^{14}\,\mathrm{Hz}$$

$$V_0 = \frac{(6.63 \times 10^{-34}\,\mathrm{J\,s}) \times (4.9 \times 10^{14}\,\mathrm{Hz})}{1.6 \times 10^{-19}\,\mathrm{C}}$$

$$V_0 = \frac{3.25 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.0\,\mathrm{V}$$

Photoelectric emission occurs only if the energy of an incident photon is at least equal to the work function $$\phi_0$$ of the metal. Equivalently, the incident wavelength must not exceed the threshold wavelength $$\lambda_0$$.

Energy of an incident photon of wavelength $$\lambda = 330\,\mathrm{nm} = 330 \times 10^{-9}\,\mathrm{m}$$:

$$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34}) \times (3.0 \times 10^{8})}{330 \times 10^{-9}}$$

$$E = \frac{1.989 \times 10^{-25}}{3.30 \times 10^{-7}} = 6.03 \times 10^{-19}\,\mathrm{J}$$

$$E = \frac{6.03 \times 10^{-19}}{1.6 \times 10^{-19}}\,\mathrm{eV} \approx 3.77\,\mathrm{eV}$$

Comparison. The work function is $$\phi_0 = 4.2\,\mathrm{eV}$$. Since the photon energy is less than the work function,

$$E = 3.77\,\mathrm{eV} \;<\; \phi_0 = 4.2\,\mathrm{eV},$$

each photon carries less energy than is needed to liberate an electron, so no photoelectric emission takes place.

As a check, the threshold wavelength of the metal is

$$\lambda_0 = \frac{hc}{\phi_0} = \frac{1.989 \times 10^{-25}}{4.2 \times 1.6 \times 10^{-19}} \approx 2.96 \times 10^{-7}\,\mathrm{m} \approx 296\,\mathrm{nm}$$

Since the incident wavelength $$330\,\mathrm{nm}$$ is greater than $$\lambda_0 \approx 296\,\mathrm{nm}$$, emission cannot occur — confirming the conclusion.

The maximum kinetic energy of the ejected electrons is

$$K_{\max} = \frac{1}{2} m v_{\max}^2$$

With $$m = 9.11 \times 10^{-31}\,\mathrm{kg}$$ and $$v_{\max} = 6.0 \times 10^{5}\,\mathrm{m/s}$$:

$$K_{\max} = \frac{1}{2} \times (9.11 \times 10^{-31}) \times (6.0 \times 10^{5})^2$$

$$K_{\max} = \frac{1}{2} \times (9.11 \times 10^{-31}) \times (3.6 \times 10^{11}) = 1.64 \times 10^{-19}\,\mathrm{J}$$

By Einstein's photoelectric equation, $$h\nu = h\nu_0 + K_{\max}$$, so the threshold frequency is

$$\nu_0 = \nu - \frac{K_{\max}}{h}$$

$$\frac{K_{\max}}{h} = \frac{1.64 \times 10^{-19}}{6.63 \times 10^{-34}} = 2.47 \times 10^{14}\,\mathrm{Hz}$$

$$\nu_0 = 7.21 \times 10^{14}\,\mathrm{Hz} - 2.47 \times 10^{14}\,\mathrm{Hz}$$

$$\nu_0 \approx 4.74 \times 10^{14}\,\mathrm{Hz}$$

The energy of a photon of the incident light ($$\lambda = 488\,\mathrm{nm} = 488 \times 10^{-9}\,\mathrm{m}$$) is

$$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34}) \times (3.0 \times 10^{8})}{488 \times 10^{-9}}$$

$$E = \frac{1.989 \times 10^{-25}}{4.88 \times 10^{-7}} = 4.08 \times 10^{-19}\,\mathrm{J}$$

$$E = \frac{4.08 \times 10^{-19}}{1.6 \times 10^{-19}}\,\mathrm{eV} \approx 2.55\,\mathrm{eV}$$

By Einstein's photoelectric equation, the photon energy is shared between the work function and the maximum kinetic energy of the photoelectrons, the latter equal to $$eV_0$$:

$$E = \phi_0 + eV_0 \quad\Rightarrow\quad \phi_0 = E - eV_0$$

With a stopping potential $$V_0 = 0.38\,\mathrm{V}$$, we have $$eV_0 = 0.38\,\mathrm{eV}$$, so

$$\phi_0 = 2.55\,\mathrm{eV} - 0.38\,\mathrm{eV} = 2.17\,\mathrm{eV}$$

Consider electromagnetic radiation of wavelength $$\lambda$$. Since it travels at speed $$c$$, its frequency is $$\nu = c/\lambda$$.

Energy of its quantum (photon). By Planck's relation,

$$E = h\nu = \frac{hc}{\lambda}$$

Momentum of the photon. A photon is massless and travels at speed $$c$$; from relativity its energy and momentum are related by $$E = pc$$. Hence

$$p = \frac{E}{c} = \frac{hc/\lambda}{c} = \frac{h}{\lambda}$$

de Broglie wavelength of the photon. The de Broglie wavelength of any particle carrying momentum $$p$$ is $$\lambda_{dB} = h/p$$. Substituting the photon momentum found above,

$$\lambda_{dB} = \frac{h}{p} = \frac{h}{\,h/\lambda\,} = \lambda$$

Thus the de Broglie wavelength of a photon is exactly equal to the wavelength of the electromagnetic radiation of which it is the quantum.

$$\lambda_{dB} = \frac{h}{p} = \lambda$$