📢 Too many exams? Don’t know which one suits you best? Book Your Free Expert 👉 call Now!
  • google app store apple app store
    • ✖

      Question

      The stopping potential for photoelectrons emitted from

      a metal surface when light of frequency 8×1014 Hz is incident is 3 V. What is the work function of the metal?
      A 0.309 eV Correct Answer Incorrect Answer
      B 0.246 eV Correct Answer Incorrect Answer
      C 0.157 eV Correct Answer Incorrect Answer
      D 0.758 eV Correct Answer Incorrect Answer

      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.

      Practice Next
      More Physics GS Questions

      Relevant for Exams:

      ask-question