A student is studying the photodissociation (dissociation with light) of I2 into 2I. When a sample of I2 is irradiated with a power of 255 mW at 590 nm for 35 seconds, 0.0256 mmol of I forms. Assuming complete absorption of the incident radiation, what is the quantum yield, f, of the reaction? (See Problem 93 for definition of quantum yield.)

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1. Identify the given values: power (P) = 255 mW, wavelength (\(\lambda\)) = 590 nm, time (t) = 35 s, moles of I formed = 0.0256 mmol.
2. Convert the power from milliwatts to watts: 1 mW = 0.001 W, so 255 mW = 0.255 W.
3. Calculate the energy of a single photon using the formula \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant (6.626 \times 10^{-34} \text{ J s}) and \(c\) is the speed of light (3.00 \times 10^8 \text{ m/s}). Convert \(\lambda\) from nm to meters.
4. Calculate the total energy absorbed by the sample using the formula \(\text{Total Energy} = P \times t\).
5. Determine the number of photons absorbed using the formula \(\text{Number of photons} = \frac{\text{Total Energy}}{E}\), and then calculate the quantum yield \(f\) using \(f = \frac{\text{moles of I formed}}{\text{Number of photons absorbed}}\).