Practice question
Question
A photon has a momentum of \( 2.0 \times 10^{-27} \, \text{kg m/s} \). What is its energy in joules?
(Take \( c = 3 \times 10^8 \, \text{m/s} \))
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. For a photon, p = (E/c) . E = p c = 2.0 × 10⁻²⁷ × 3 × 10⁸ = 6.0 × 10⁻¹⁹ J . Applying E = h f = h c/λ, p = h/λ, K_max = h
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