Question

The work function of a metal is $2\text{ eV}$. Photoelectric emission will occur when the metal is illuminated with light of energy:

• A. $1\text{ eV}$
• B. $1.5\text{ eV}$
• C. $2.5\text{ eV}$
• D. $2\text{ eV}$ only

Hint:

Think of the work function as an escape toll. If the toll is 2, any incoming photon with less than 2 will fail. Only an option strictly equal to or higher than the toll—like 2.5 eV—can trigger emission.

Answer 1

The Correct Option is C

Concept: According to Einstein's Photoelectric Equation, photoelectric emission is an instantaneous, one-to-one interaction between an incoming photon and a bound electron. The energy ($E$) of the incident photon must satisfy: \[ E = \phi_0 + K_{\text{max}} \] Where:
• $\phi_0$ is the work function of the metal (the minimum threshold energy required to just liberate an electron from the metal surface).
• $K_{\text{max}}$ is the maximum kinetic energy of the ejected photoelectron.

Step 1: Establish the conditional rule for electron emission.
For an electron to successfully escape the metal surface, the energy of the incident photon ($E$) must be greater than or equal to the metal's work function ($\phi_0$): \[ E \ge \phi_0 \]
• If $E & Lt; \phi_0$: The photon doesn't have enough energy to break the electron's atomic bonds, so no emission occurs, regardless of light intensity or duration.
• If $E \ge \phi_0$: Electron emission happens instantly, and any leftover energy becomes kinetic energy ($K_{\text{max}} = E - \phi_0$).

Step 2: Evaluate the given energy options against the threshold.
The metal's work function is given as $\phi_0 = 2\text{ eV}$. Let us check each option:
• Option (A): $1\text{ eV} & Lt; 2\text{ eV}$ $\rightarrow$ No emission.
• Option (B): $1.5\text{ eV} & Lt; 2\text{ eV}$ $\rightarrow$ No emission.
• Option (C): $2.5\text{ eV} > 2\text{ eV}$ $\rightarrow$ Emission occurs successfully, and the electron retains $0.5\text{ eV}$ of kinetic energy.