Question

Radiation of wavelength $\lambda$ is incident on a photosensitive surface. Find the de Broglie wavelength of electrons emitted from the surface. Assume that the work function of the surface is negligible.

Hint:

Notice that $\lambda_{dB} \propto \sqrt{\lambda}$. Longer incident light wavelength means lower photon energy, resulting in lower electron momentum and hence a longer de Broglie wavelength for the emitted electron.

Answer 1

Concept:
• Energy of an incident photon of wavelength $\lambda$ is $E = \frac{hc}{\lambda}$.

• Einstein's photoelectric equation states $K_{max} = E - \Phi_0$. For negligible work function ($\Phi_0 \approx 0$), maximum kinetic energy of emitted electron is $K_{max} = E = \frac{hc}{\lambda}$.

• The de Broglie wavelength of an electron of mass $m_e$ and kinetic energy $K$ is $\lambda_{dB} = \frac{h}{\sqrt{2 m_e K}}$.

Step 1: Determine maximum kinetic energy of photoelectrons
Given incident radiation wavelength = $\lambda$.
Work function of photosensitive surface $\Phi_0 \approx 0$.
Maximum kinetic energy $K$ acquired by emitted electrons:
\[ K = \frac{hc}{\lambda} \]

Step 2: Express de Broglie wavelength in terms of kinetic energy
The de Broglie wavelength $\lambda_{dB}$ of the photoelectrons is:
\[ \lambda_{dB} = \frac{h}{\sqrt{2 m_e K}} \]

Step 3: Substitute kinetic energy into the de Broglie relation
Substitute $K = \frac{hc}{\lambda}$:
\[ \lambda_{dB} = \frac{h}{\sqrt{2 m_e \left(\frac{hc}{\lambda}\right)}} \]
\[ \lambda_{dB} = \frac{h}{\sqrt{\frac{2 m_e h c}{\lambda}}} = \sqrt{\frac{h^2 \lambda}{2 m_e h c}} \]
Simplifying $h$ inside the square root:
\[ \lambda_{dB} = \sqrt{\frac{h \lambda}{2 m_e c}} \]

Step 4: Conclusion
The de Broglie wavelength of the most energetic photoelectrons emitted from the surface is $\lambda_{dB} = \sqrt{\frac{h \lambda}{2 m_e c}}$.