What is the energy of a photon with a frequency of 6 x 10^14 Hz?
Correct Answer: 4.14 x 10^-19 J
- Step 1: Identify the formula for the energy of a photon, which is E = h * f.
- Step 2: Note the value of Planck's constant (h), which is 6.63 x 10^-34 J·s.
- Step 3: Identify the frequency (f) of the photon, which is given as 6 x 10^14 Hz.
- Step 4: Substitute the values into the formula: E = (6.63 x 10^-34 J·s) * (6 x 10^14 Hz).
- Step 5: Perform the multiplication: 6.63 * 6 = 39.78 and 10^-34 * 10^14 = 10^-20.
- Step 6: Combine the results: E = 39.78 x 10^-20 J.
- Step 7: Convert 39.78 x 10^-20 J to scientific notation: E = 4.14 x 10^-19 J.
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