Question

A small metal plate (work function \(\phi\)) is kept at a distance \(d\) from a single ionized fixed ion. A monochromatic light beam is incident on the metal plate, and photoelectrons are emitted. What is the maximum wavelength of the light beam, so that some of the photoelectrons may go around the ion along a circle?

• A. \(\dfrac{8\pi\epsilon_0 dhc}{e^2 + 8\pi\epsilon_0\phi d}\)
• B. \(\dfrac{8\pi\epsilon_0 dhc}{e^2 - 8\pi\epsilon_0\phi d}\)
• C. \(\dfrac{8\pi\epsilon_0 dhc + e^2}{8\pi\epsilon_0\phi d}\)
• D. \(\dfrac{8\pi\epsilon_0 dhc - e^2}{8\pi\epsilon_0\phi d}\)

Hint:

Set the maximum photoelectron KE \(hc/\lambda-\phi\) equal to the orbital KE \(e^2/(8\pi\epsilon_0 d)\).

Answer 1

The Correct Option is A

Step 1: A photoelectron of the plate is emitted with maximum kinetic energy given by Einstein's equation \(K = \dfrac{hc}{\lambda} - \phi\).

Step 2: For the electron (charge \(-e\)) to move in a circle of radius \(d\) about the singly ionized ion (charge \(+e\)), the Coulomb attraction supplies the centripetal force:
\[\frac{1}{4\pi\epsilon_0}\frac{e^2}{d^2} = \frac{mv^2}{d}\]
so \(mv^2 = \dfrac{e^2}{4\pi\epsilon_0 d}\).

Step 3: The kinetic energy needed for this circular motion is
\[K_{circ} = \tfrac{1}{2}mv^2 = \frac{e^2}{8\pi\epsilon_0 d}\]

Step 4: Maximum wavelength means minimum photon energy, so the maximum available kinetic energy just equals \(K_{circ}\):
\[\frac{hc}{\lambda} - \phi = \frac{e^2}{8\pi\epsilon_0 d}\]
\[\frac{hc}{\lambda} = \phi + \frac{e^2}{8\pi\epsilon_0 d} = \frac{8\pi\epsilon_0\phi d + e^2}{8\pi\epsilon_0 d}\]

Step 5: Solving for \(\lambda\):
\[\boxed{\lambda_{max} = \dfrac{8\pi\epsilon_0 dhc}{e^2 + 8\pi\epsilon_0\phi d}}\]