Question

Find the maximum kinetic energy of the photo-electrons ejected, when light of wavelength 3300 Å is incident on a Cesium surface. Work function of Cesium = 1.9 eV.

Hint:

Memorizing the constant combination \(hc \approx 1240 \, \text{eV} \cdot \text{nm}\) or \(12400 \, \text{eV} \cdot \text{Å}\) is extremely helpful for modern physics problems. It allows you to directly convert wavelength in nanometers or Angstroms to photon energy in electron-volts without dealing with Planck's constant, the speed of light, and the charge of an electron individually.

Answer 1

Step 1: Understanding the Concept (Photoelectric Effect):
This problem uses Einstein's photoelectric equation, which describes the energy balance when a photon strikes a metal surface and ejects an electron (a photoelectron). The energy of the incident photon is used partly to overcome the work function of the metal and the rest is converted into the kinetic energy of the ejected electron.

Step 2: Key Formula or Approach:
Einstein's photoelectric equation is: \[ K.E._{max} = E_{photon} - \phi_0 \] where: \begin{itemize} \item \(K.E._{max}\) is the maximum kinetic energy of the ejected photoelectron. \item \(E_{photon}\) is the energy of the incident photon. \item \(\phi_0\) is the work function of the metal. \end{itemize} The energy of a photon can be calculated from its wavelength \(\lambda\). A very useful shortcut for calculations where wavelength is in Angstroms (Å) and energy is needed in electron-volts (eV) is: \[ E_{photon} (\text{in eV}) = \frac{hc}{e\lambda} \approx \frac{12400}{\lambda (\text{in Å})} \]

Step 3: Detailed Calculation:
\begin{enumerate} \item Given Data: \begin{itemize} \item Wavelength of incident light, \(\lambda = 3300\) Å. \item Work function of Cesium, \(\phi_0 = 1.9\) eV. \end{itemize} \item Calculate Photon Energy (\(E_{photon}\)): Using the shortcut formula: \[ E_{photon} = \frac{12400}{\lambda (\text{in Å})} \, \text{eV} = \frac{12400}{3300} \, \text{eV} \] \[ E_{photon} = \frac{124}{33} \, \text{eV} \approx 3.757 \, \text{eV} \] Let's use \(E_{photon} \approx 3.76\) eV for our calculation. \item Calculate Maximum Kinetic Energy (\(K.E._{max}\)): Now, substitute the values into the photoelectric equation: \[ K.E._{max} = E_{photon} - \phi_0 \] \[ K.E._{max} = 3.76 \, \text{eV} - 1.9 \, \text{eV} \] \[ K.E._{max} = 1.86 \, \text{eV} \] \end{enumerate}

Step 4: Final Answer:
The maximum kinetic energy of the ejected photo-electrons is 1.86 eV.