A metal gives (frequency in 10¹⁴ Hz, stopping potential in V) readings (5.5, 0.18), (6, 0.39), (6.5, 0.59), (7, 0.8), (7.5, 1.01). Use e = 1.60 × 10⁻¹⁹ C and mₑ = 9.11 × 10⁻³¹ kg. (a) Find gradient and h. (b) Estimate threshold frequency and work function in eV. (c) At 7.0 × 10¹⁴ Hz find maximum speed. (d) Explain intensity below/above threshold. (e) Explain photon evidence.

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Gradient 4.15 × 10⁻¹⁵ V s; h = 6.64 × 10⁻³⁴ J s; threshold 5.07 × 10¹⁴ Hz; work function 2.10 eV; maximum speed 5.30 × 10⁵ m/s. Endpoint gradient = (1.01 − 0.18)/[(7.5 − 5.5) × 10¹⁴] = 4.15 × 10⁻¹⁵ V s. From eVs = hf − φ, gradient = h/e. Thus h = 1.60 × 10⁻¹⁹ × 4.15 × 10⁻¹⁵ = 6.64 × 10⁻³⁴ J s. Threshold f₀ = 5.5 × 10¹⁴ − 0.18/(4.15 × 10⁻¹⁵) = 5.07 × 10¹⁴ Hz. Work function is gradient × f₀ = 2.10 eV. At 7.0 × 10¹⁴ Hz, Vs = 0.80 V and Kmax = eVs. Then v = √(2eVs/m) = 5.30 × 10⁵ m/s. Below threshold, more intensity gives no ideal emission; above it, intensity raises electron rate but not maximum energy. Immediate emission, threshold and linear energy–frequency relation support photons hf.

Photoelectric stopping-potential evidence

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Photoelectric stopping potential

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Photoelectric stopping potential — supplied values
Frequency / 10¹⁴ Hz	Stopping potential / V
5.5	0.18
6	0.39
6.5	0.59
7	0.8
7.5	1.01