Dual Nature of Radiation and Matter

Given: Accelerating voltage, $V = 30\,\mathrm{kV} = 30 \times 10^3\,\mathrm{V}$

Concept: The maximum energy of the X-ray photon equals the kinetic energy gained by the electron:
$$eV = h\nu_{\max} = \frac{hc}{\lambda_{\min}}$$

(a) Maximum frequency:
$$\nu_{\max} = \frac{eV}{h} = \frac{1.6 \times 10^{-19}\,\mathrm{C} \times 30 \times 10^3\,\mathrm{V}}{6.63 \times 10^{-34}\,\mathrm{J\,s}}$$
$$\nu_{\max} = \frac{4.8 \times 10^{-15}}{6.63 \times 10^{-34}} = 7.24 \times 10^{18}\,\mathrm{Hz}$$

(b) Minimum wavelength:
$$\lambda_{\min} = \frac{c}{\nu_{\max}} = \frac{3 \times 10^8\,\mathrm{m/s}}{7.24 \times 10^{18}\,\mathrm{Hz}}$$
$$\boxed{\lambda_{\min} = 0.0414\,\mathrm{nm} \approx 4.14 \times 10^{-11}\,\mathrm{m}}$$

Given:
- Work function, $\phi_0 = 2.14\,\mathrm{eV} = 2.14 \times 1.6 \times 10^{-19}\,\mathrm{J} = 3.424 \times 10^{-19}\,\mathrm{J}$
- Frequency of incident light, $\nu = 6 \times 10^{14}\,\mathrm{Hz}$

Energy of incident photon:
$$E = h\nu = 6.63 \times 10^{-34} \times 6 \times 10^{14} = 3.978 \times 10^{-19}\,\mathrm{J} = 2.486\,\mathrm{eV}$$

(a) Maximum kinetic energy:
Using Einstein's photoelectric equation:
$$K_{\max} = h\nu - \phi_0 = 3.978 \times 10^{-19} - 3.424 \times 10^{-19}$$
$$\boxed{K_{\max} = 0.554 \times 10^{-19}\,\mathrm{J} \approx 0.34\,\mathrm{eV}}$$

(b) Stopping potential:
$$eV_0 = K_{\max}$$
$$V_0 = \frac{K_{\max}}{e} = \frac{0.554 \times 10^{-19}}{1.6 \times 10^{-19}}$$
$$\boxed{V_0 = 0.34\,\mathrm{V}}$$

(c) Maximum speed of emitted photoelectrons:
$$\frac{1}{2}m v_{\max}^2 = K_{\max}$$
$$v_{\max} = \sqrt{\frac{2K_{\max}}{m}} = \sqrt{\frac{2 \times 0.554 \times 10^{-19}}{9.11 \times 10^{-31}}}$$
$$v_{\max} = \sqrt{1.217 \times 10^{11}} = \sqrt{12.17 \times 10^{10}}$$
$$\boxed{v_{\max} \approx 3.49 \times 10^5\,\mathrm{m/s}}$$

Given: Stopping (cut-off) potential, $V_0 = 1.5\,\mathrm{V}$

Concept: The stopping potential is related to the maximum kinetic energy by:
$$K_{\max} = eV_0$$

Calculation:
$$K_{\max} = eV_0 = 1.6 \times 10^{-19}\,\mathrm{C} \times 1.5\,\mathrm{V}$$
$$\boxed{K_{\max} = 2.4 \times 10^{-19}\,\mathrm{J} = 1.5\,\mathrm{eV}}$$

Given:
- Wavelength, $\lambda = 632.8\,\mathrm{nm} = 632.8 \times 10^{-9}\,\mathrm{m}$
- Power, $P = 9.42\,\mathrm{mW} = 9.42 \times 10^{-3}\,\mathrm{W}$

(a) Energy and momentum of each photon:

Energy:
$$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{632.8 \times 10^{-9}}$$
$$\boxed{E = 3.14 \times 10^{-19}\,\mathrm{J} \approx 1.96\,\mathrm{eV}}$$

Momentum:
$$p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{632.8 \times 10^{-9}}$$
$$\boxed{p = 1.05 \times 10^{-27}\,\mathrm{kg\,m/s}}$$

(b) Number of photons per second:

Let $n$ be the number of photons emitted per second. Then:
$$P = nE \implies n = \frac{P}{E}$$
$$n = \frac{9.42 \times 10^{-3}}{3.14 \times 10^{-19}}$$
$$\boxed{n = 3.0 \times 10^{16}\,\mathrm{photons/s}}$$

(c) Speed of hydrogen atom with same momentum:

Mass of hydrogen atom, $m_H = 1.67 \times 10^{-27}\,\mathrm{kg}$

$$p = m_H v \implies v = \frac{p}{m_H} = \frac{1.05 \times 10^{-27}}{1.67 \times 10^{-27}}$$
$$\boxed{v = 0.629\,\mathrm{m/s} \approx 0.63\,\mathrm{m/s}}$$

Given: Slope of $V_0$ vs $\nu$ graph $= 4.12 \times 10^{-15}\,\mathrm{V\,s}$

Concept: From Einstein's photoelectric equation:
$$eV_0 = h\nu - \phi_0$$
$$V_0 = \frac{h}{e}\nu - \frac{\phi_0}{e}$$

The slope of $V_0$ vs $\nu$ is $\dfrac{h}{e}$.

Calculation:
$$\frac{h}{e} = 4.12 \times 10^{-15}\,\mathrm{V\,s}$$
$$h = 4.12 \times 10^{-15} \times e = 4.12 \times 10^{-15} \times 1.6 \times 10^{-19}$$
$$\boxed{h = 6.592 \times 10^{-34}\,\mathrm{J\,s} \approx 6.59 \times 10^{-34}\,\mathrm{J\,s}}$$

This is in close agreement with the standard value of Planck's constant $h = 6.63 \times 10^{-34}\,\mathrm{J\,s}$.

Given:
- Threshold frequency, $\nu_0 = 3.3 \times 10^{14}\,\mathrm{Hz}$
- Incident frequency, $\nu = 8.2 \times 10^{14}\,\mathrm{Hz}$

Concept: Einstein's photoelectric equation:
$$eV_0 = h(\nu - \nu_0)$$
$$V_0 = \frac{h(\nu - \nu_0)}{e}$$

Calculation:
$$V_0 = \frac{6.63 \times 10^{-34} \times (8.2 - 3.3) \times 10^{14}}{1.6 \times 10^{-19}}$$
$$V_0 = \frac{6.63 \times 10^{-34} \times 4.9 \times 10^{14}}{1.6 \times 10^{-19}}$$
$$V_0 = \frac{32.487 \times 10^{-20}}{1.6 \times 10^{-19}} = \frac{3.2487 \times 10^{-19}}{1.6 \times 10^{-19}}$$
$$\boxed{V_0 = 2.03\,\mathrm{V}}$$

Given:
- Work function, $\phi_0 = 4.2\,\mathrm{eV}$
- Wavelength of incident radiation, $\lambda = 330\,\mathrm{nm} = 330 \times 10^{-9}\,\mathrm{m}$

Energy of incident photon:
$$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{330 \times 10^{-9}}$$
$$E = \frac{19.89 \times 10^{-26}}{330 \times 10^{-9}} = 6.027 \times 10^{-19}\,\mathrm{J}$$
$$E = \frac{6.027 \times 10^{-19}}{1.6 \times 10^{-19}}\,\mathrm{eV} = 3.77\,\mathrm{eV}$$

Comparison:
The energy of the incident photon ($E = 3.77\,\mathrm{eV}$) is less than the work function ($\phi_0 = 4.2\,\mathrm{eV}$).

E < \phi_0

Conclusion: No, photoelectric emission will not occur for this metal with the given incident radiation, since the photon energy is insufficient to overcome the work function.

Given:
- Incident frequency, $\nu = 7.21 \times 10^{14}\,\mathrm{Hz}$
- Maximum speed of ejected electrons, $v_{\max} = 6.0 \times 10^5\,\mathrm{m/s}$
- Mass of electron, $m = 9.11 \times 10^{-31}\,\mathrm{kg}$

Concept: Einstein's photoelectric equation:
$$h\nu = \phi_0 + \frac{1}{2}mv_{\max}^2 = h\nu_0 + \frac{1}{2}mv_{\max}^2$$

Maximum kinetic energy:
$$K_{\max} = \frac{1}{2}mv_{\max}^2 = \frac{1}{2} \times 9.11 \times 10^{-31} \times (6.0 \times 10^5)^2$$
$$K_{\max} = \frac{1}{2} \times 9.11 \times 10^{-31} \times 3.6 \times 10^{11}$$
$$K_{\max} = 1.64 \times 10^{-19}\,\mathrm{J}$$

Threshold frequency:
$$h\nu_0 = h\nu - K_{\max}$$
$$\nu_0 = \nu - \frac{K_{\max}}{h} = 7.21 \times 10^{14} - \frac{1.64 \times 10^{-19}}{6.63 \times 10^{-34}}$$
$$\nu_0 = 7.21 \times 10^{14} - 2.474 \times 10^{14}$$
$$\boxed{\nu_0 = 4.74 \times 10^{14}\,\mathrm{Hz}}$$

Given:
- Wavelength of incident light, $\lambda = 488\,\mathrm{nm} = 488 \times 10^{-9}\,\mathrm{m}$
- Stopping potential, $V_0 = 0.38\,\mathrm{V}$

Energy of incident photon:
$$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}$$
$$E = \frac{19.89 \times 10^{-26}}{488 \times 10^{-9}} = 4.075 \times 10^{-19}\,\mathrm{J} = 2.547\,\mathrm{eV}$$

Work function using Einstein's equation:
$$\phi_0 = E - eV_0 = 2.547\,\mathrm{eV} - 0.38\,\mathrm{eV}$$
$$\boxed{\phi_0 = 2.17\,\mathrm{eV}}$$

Concept: de Broglie wavelength: $\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$

(a) Bullet:
- $m = 0.040\,\mathrm{kg}$, $v = 1.0\,\mathrm{km/s} = 1.0 \times 10^3\,\mathrm{m/s}$
$$p = mv = 0.040 \times 1.0 \times 10^3 = 40\,\mathrm{kg\,m/s}$$
$$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{40}$$
$$\boxed{\lambda = 1.66 \times 10^{-35}\,\mathrm{m}}$$

(b) Ball:
- $m = 0.060\,\mathrm{kg}$, $v = 1.0\,\mathrm{m/s}$
$$p = mv = 0.060 \times 1.0 = 0.060\,\mathrm{kg\,m/s}$$
$$\lambda = \frac{6.63 \times 10^{-34}}{0.060}$$
$$\boxed{\lambda = 1.105 \times 10^{-32}\,\mathrm{m} \approx 1.1 \times 10^{-32}\,\mathrm{m}}$$

(c) Dust particle:
- $m = 1.0 \times 10^{-9}\,\mathrm{kg}$, $v = 2.2\,\mathrm{m/s}$
$$p = mv = 1.0 \times 10^{-9} \times 2.2 = 2.2 \times 10^{-9}\,\mathrm{kg\,m/s}$$
$$\lambda = \frac{6.63 \times 10^{-34}}{2.2 \times 10^{-9}}$$
$$\boxed{\lambda = 3.01 \times 10^{-25}\,\mathrm{m} \approx 3.0 \times 10^{-25}\,\mathrm{m}}$$

In all three cases, the de Broglie wavelengths are extremely small and far beyond experimental measurement, which is why macroscopic objects do not exhibit wave-like properties.

To Show: $\lambda_{\text{EM}} = \lambda_{\text{de Broglie}}$ for a photon.

For electromagnetic radiation:
The wavelength of electromagnetic radiation is $\lambda$, and the energy of a photon is:
$$E = h\nu = \frac{hc}{\lambda}$$

Momentum of a photon:
From Einstein's mass-energy relation and the fact that a photon has zero rest mass, the momentum of a photon is:
$$p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda}$$

de Broglie wavelength of the photon:
Using de Broglie's relation $\lambda_{\text{dB}} = \dfrac{h}{p}$:
$$\lambda_{\text{dB}} = \frac{h}{p} = \frac{h}{h/\lambda} = \lambda$$

Conclusion:
$$\boxed{\lambda_{\text{dB}} = \lambda}$$

Thus, the de Broglie wavelength of a photon is equal to the wavelength of the corresponding electromagnetic radiation. This result shows the inherent consistency between the wave and particle descriptions of electromagnetic radiation.