Question

In a photoelectric experiment, the emitter plate is irradiated with radiation of \(200\ \text{nm}\). The photocurrent becomes zero when the collector plate potential is \(-0.80\ \text{V}\). Calculate the work function (in eV) of the emitter.

Hint:

For photoelectric effect problems, remember the two important formulas: \[ E_{\text{photon}}=\frac{1240}{\lambda(\text{nm})}\ \text{eV} \] and \[ h\nu=\phi+eV_0. \] A quick method is: \[ \boxed{\phi=\frac{1240}{\lambda(\text{nm})}-V_0} \] when all quantities are expressed in electron volts and volts.

Answer 1

Concept: The photoelectric effect is based on Einstein's photoelectric equation, which states that when light of sufficiently high frequency falls on a metal surface, electrons are emitted from the surface. The energy of the incident photon is used in two ways:
• A part of the energy is used to overcome the work function of the metal.
• The remaining energy appears as the maximum kinetic energy of the emitted photoelectrons. Mathematically, \[ h\nu=\phi+K_{\max} \] where, \[ h\nu=\text{energy of the incident photon}, \] \[ \phi=\text{work function of the metal}, \] and \[ K_{\max}=\text{maximum kinetic energy of the emitted photoelectrons}. \] The maximum kinetic energy is related to the stopping potential \(V_0\) by \[ K_{\max}=eV_0. \] Thus, once the photon energy and stopping potential are known, the work function can be calculated using Einstein's photoelectric equation.

Step 1: Calculate the energy of the incident photon.
The wavelength of the incident radiation is \[ \lambda=200\ \text{nm}. \] The energy of a photon in electron volt can be calculated directly using \[ E=\frac{1240}{\lambda(\text{in nm})}\ \text{eV}. \] Substituting the given value, \[ E=\frac{1240}{200} \] \[ E=6.2\ \text{eV}. \] Therefore, the energy of each incident photon is \[ h\nu=6.2\ \text{eV}. \]

Step 2: Determine the maximum kinetic energy of the emitted photoelectrons.
The photocurrent becomes zero when the collector plate potential is \[ V_0=0.80\ \text{V}. \] Hence, \[ K_{\max}=eV_0. \] In electron volt, \[ K_{\max}=0.80\ \text{eV}. \]

Step 3: Apply Einstein's photoelectric equation to calculate the work function.
Using \[ h\nu=\phi+K_{\max}, \] we get \[ \phi=h\nu-K_{\max}. \] Substituting the values, \[ \phi=6.2-0.8 \] \[ \phi=5.4\ \text{eV}. \] Therefore, the work function of the emitter is \[ \boxed{\phi=5.4\ \text{eV}} \]