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Question

The threshold frequency of a metal is \( 6.0 \times 10^{14} \, \text{Hz} \). What is its work function
in eV? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( 1 \, \text{eV} = 1.6 \times 10^{-19} \,
\text{J} \))

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Explanation

**Photon energy** E = h f = h c/λ, momentum p = h/λ = E/c, power P = N E/t, N number of photons per second, wavelength λ = h c/E, for photon momentum 1.5×10⁻²⁷ kg·m/s frequency f = p c/h? Actually E = p c, f = E/h = p c/h =1.5×10⁻²⁷×3×10⁸/6.63×10⁻³⁴=6.79×10¹⁴ Hz, energy E= h f =4.5×10⁻¹⁹ J, for 5.0×10⁻¹⁹ J frequency f=E/h=7.54×10¹⁴ Hz. Φ₀ = h v₀ = 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴ = 3.978 × 10⁻¹⁹ J . Φ₀ = (3.978 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 2.486 eV . Applying E = h f = h c/λ, p =

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