Given that the energy of a photon is 4.38 × 10^{-18} J, what is its frequency (s^{-1})? (Planck's constant h = 6.626 × 10^{-34} J·s)

• A. 1.48 × 10^{16} s^{-1}
• B. 2.92 × 10^{14} s^{-1}
• C. 6.61 × 10^{15} s^{-1}
• D. 4.38 × 10^{-18} s^{-1}

Answer: (C)

Verified step by step guidance

1. Recall the relationship between the energy of a photon and its frequency, given by the equation \(E = h \nu\), where \(E\) is the energy, \(h\) is Planck's constant, and \(\nu\) is the frequency.
2. Identify the known values from the problem: \(E = 4.38 \times 10^{-18}\) J and \(h = 6.626 \times 10^{-34}\) J\cdot s.
3. Rearrange the equation to solve for frequency \(\nu\): \(\nu = \frac{E}{h}\).
4. Substitute the known values into the rearranged equation: \(\nu = \frac{4.38 \times 10^{-18}}{6.626 \times 10^{-34}}\).
5. Perform the division to find the frequency \(\nu\) in s\(^{-1}\) (Hz), which will give the frequency of the photon.