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: Work energywise. The photon delivers energy $E_{ph} = hc/\lambda$. Part of it, $\phi$, frees the electron; the rest becomes the electron's kinetic energy $K = hc/\lambda - \phi$.

Step 2: A stable circular orbit of radius $d$ around a charge $+e$ requires a definite speed. Balancing the electrostatic pull against the centripetal requirement, $\dfrac{e^2}{4\pi\epsilon_0 d^2} = \dfrac{mv^2}{d}$, gives the orbital kinetic energy $\dfrac{1}{2}mv^2 = \dfrac{e^2}{8\pi\epsilon_0 d}$.

Step 3: To just make circular motion possible, the largest kinetic energy the plate can supply must reach this value:
\[\frac{hc}{\lambda} - \phi = \frac{e^2}{8\pi\epsilon_0 d}\]

Step 4: Combine the two terms on the right over the common denominator $8\pi\epsilon_0 d$:
\[\frac{hc}{\lambda} = \frac{e^2 + 8\pi\epsilon_0\phi d}{8\pi\epsilon_0 d}\]

Step 5: Inverting and multiplying by $hc$ gives the largest permitted wavelength:
\[\boxed{\lambda_{max} = \dfrac{8\pi\epsilon_0 dhc}{e^2 + 8\pi\epsilon_0\phi d}}\]