Question

A gamma ray with an energy of 3.40 × 10-14 joules strikes a photographic plate. We know that Planck's constant is 6.63 × 10-34 joule·seconds. What is the frequency of the photon?

Answer 1

Answer on Question #65203 – Physics/Molecular physics - Thermodynamics

Question: A gamma ray with an energy of $3.40 \times 10^{-14}$ joules strikes a photographic plate. We know that Planck's constant is $6.63 \times 10^{-34}$ joule seconds. What is the frequency of the photon?

Solution: The Planck's constant $(h)$ links a photon's energy $(E)$ and photon's frequency $(f)$ by equation: $E = hf$, so $f = E / h = \frac{3.40 \times 10^{-14} J}{6.63 \times 10^{-34} J \times s} = 0.513 \times 10^{20} Hz$.

Answer: The frequency of the photon is $0.513 \times 10^{20} \mathrm{~Hz}$.

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