Question

The work function of a metal is 5 eV. What is the kinetic energy of the photoelectron ejected from the metal surface if the energy of the incident radiation is 6.2 eV? ($1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$)

• A. $6.626 \times 10^{-19} \text{ J}$
• B. $8.01 \times 10^{-19} \text{ J}$
• C. $1.92 \times 10^{-18} \text{ J}$
• D. $8.010 \times 10^{-18} \text{ J}$
• E. $1.92 \times 10^{-19} \text{ J}$

Hint:

If the incident energy is less than the work function ($E < \Phi$), no photoelectrons will be ejected, regardless of the intensity of the light!

Answer 1

Answer: (E)

Understanding the Concept: The photoelectric effect is described by Einstein's photoelectric equation, which is based on the principle of conservation of energy.
• Einstein's Equation: $E_{incident} = \Phi + K_{max}$, where $\Phi$ is the work function and $K_{max}$ is the maximum kinetic energy of the emitted photoelectron.
• Unit Conversion: Kinetic energy in eV must be multiplied by $1.6 \times 10^{-19}$ to convert it to Joules (J).

Step 1: Calculate the kinetic energy in electron-volts (eV). Given incident energy $E = 6.2 \text{ eV}$ and work function $\Phi = 5 \text{ eV}$. \[ K_{max} = E - \Phi \] \[ K_{max} = 6.2 \text{ eV} - 5 \text{ eV} = 1.2 \text{ eV} \]

Step 2: Convert the kinetic energy to Joules. Using the conversion factor $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$: \[ K_{max} = 1.2 \times (1.6 \times 10^{-19} \text{ J}) \] \[ K_{max} = 1.92 \times 10^{-19} \text{ J} \]