What is the maximum kinetic energy of photoelectrons emitted if the incident light has a frequency of 8 x 10^14 Hz and the work function is 3 eV?

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Q: What is the maximum kinetic energy of photoelectrons emitted if the incident light has a frequency of 8 x 10^14 Hz and the work function is 3 eV?

Solution: Maximum kinetic energy (K.E.) = hf - Φ = (6.626 x 10^-34 J·s)(8 x 10^14 Hz) - 3 eV = 5 eV.

Steps: 9

Step 1: Identify the frequency of the incident light, which is given as 8 x 10^14 Hz.
Step 2: Identify the work function (Φ), which is given as 3 eV.
Step 3: Convert the work function from eV to Joules. (1 eV = 1.602 x 10^-19 J, so 3 eV = 3 x 1.602 x 10^-19 J).
Step 4: Calculate the energy of the incident light using the formula E = hf, where h (Planck's constant) is 6.626 x 10^-34 J·s.
Step 5: Substitute the values into the formula: E = (6.626 x 10^-34 J·s)(8 x 10^14 Hz).
Step 6: Calculate the energy of the incident light (E).
Step 7: Use the formula for maximum kinetic energy of photoelectrons: K.E. = hf - Φ.
Step 8: Substitute the calculated energy of the incident light and the work function into the K.E. formula.
Step 9: Calculate the maximum kinetic energy (K.E.).