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    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.

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