The stopping potential for photoelectrons emitted from

Solution

Given:

• Frequency of incident light: f = 8 × 10¹⁴ Hz
• Stopping potential: V₀ = 3 V

Using the photoelectric equation. Einstein's photoelectric equation states: hf = Φ + KEmax Where:

• h is Planck's constant (6.626 × 10⁻³⁴ J·s)
• f is the frequency of incident light
• Φ is the work function we need to find
• KEmax is the maximum kinetic energy of ejected electrons

The stopping potential V₀ is the potential difference needed to stop the most energetic electrons: KEmax = eV₀ Where e is the elementary charge (1.602 × 10⁻¹⁹ C). Substitute KEmax into the photoelectric equation. hf = Φ + eV₀ Φ = hf - eV₀ Calculating the work function. Φ = (6.626 × 10⁻³⁴ J·s)(8 × 10¹⁴ Hz) - (1.602 × 10⁻¹⁹ C)(3 V) Φ = 5.3008 × 10⁻¹⁹ J - 4.806 × 10⁻¹⁹ J Φ = 0.4948 × 10⁻¹⁹ J Converting to electron volts (eV) for convenience. Φ(eV) = Φ(J) / e Φ(eV) = 0.4948 × 10⁻¹⁹ J / (1.602 × 10⁻¹⁹ C) Φ(eV) = 0.309 eV Therefore, the work function of the metal is 0.309 eV.