Gold is illuminated by light of frequency [math: 1.88×1015 Hz] during a demonstration.Photoelectrons are ejected during this demonstration.If gold has a work function of [math: 7.71 × 10−19 J], determine the maximum kinetic energy of the ejected photoelectrons in Joules.Give your answer in standard form to 2 decimal places without units.For example, [math: 2.34 × 10−15] is written as 2.34e-15

Question

Gold is illuminated by light of frequency [math: 1.88×1015 Hz] during a demonstration.Photoelectrons are ejected during this demonstration.If gold has a work function of [math: 7.71 × 10−19 J], determine the maximum kinetic energy of the ejected photoelectrons in Joules.Give your answer in standard form to 2 decimal places without units.For example, [math: 2.34 × 10−15] is written as 2.34e-15

BlackTom AI · Accepted Answer

Using the photoelectric relation KE_max = hf − φ, where hf is the photon energy and φ is the work function.

1) Relevant constants and concepts
- Planck’s constant h ≈ 6.63×10−34 J·s.
- Photon energy E_photon = hf.
- Maximum kinetic energy of ejected electrons KE_max = E_photon − φ (if positive; otherwise no emission).

2) Step-by-step calculation
- Given frequency f = 1.88×10^15 Hz and φ = 7.71×10−19 J.
- Compute photon energy: E_photon = h f = (6.63×10−34 J·s)(1.88×10^15 s−1).
  Multiply mantissas: 6.63 × 1.88 = 12.4644.
  Combine powers of ten: 10^(−34+15) = 10^−19.
  So E_photon = 12.4644 × 10^−19 J = 1.24644 × 10^−18 J.
- Compute KE_max: KE_max = E_photon − φ
  Convert φ to the same power of ten: φ = 7.71 × 10^−19 J = 0.771 × 10^−18 J.
  Subtract: 1.24644×10^−18 − 0.771×10^−18 = 0.47544×10^−18 J.
  Convert to standard scientific form: 0.47544×10^−18 = 4.7544×10^−19 J.

3) Rounding and final formatting
- Round to two decimal places in standard form: 4.7544×10^−19 → 4.75×10^−19.
- Written without units as requested: 4.75e-19.

4) Why this is correct
- The photoelectric equation directly gives the maximum kinetic energy by subtracting the material’s work function from the photon energy. All steps used consistent units (Joules) and proper handling of powers of ten.

5) Common incorrect approaches
- Using an incorrect value for h, failing to align powers of ten, or forgetting to subtract the work function would give wrong results. If E_photon were smaller than φ, KE would be negative and no electrons would be emitted.

Conclusion: The maximum kinetic energy is 4.75e-19, matching the provided answer.

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