What is the work function of a metal if the threshold frequency is 5 x 10^14 Hz?

What is the work function of a metal if the threshold frequency is 5 x 10^14 Hz?

Step 1: Understand that the work function (φ) of a metal is the minimum energy needed to remove an electron from the metal.
Step 2: Know that the work function can be calculated using the formula φ = hf, where 'h' is Planck's constant and 'f' is the threshold frequency.
Step 3: Identify the value of Planck's constant, which is h = 6.63 x 10^-34 Joule seconds (J.s).
Step 4: Identify the given threshold frequency, which is f = 5 x 10^14 Hz.
Step 5: Substitute the values into the formula: φ = (6.63 x 10^-34 J.s) * (5 x 10^14 Hz).
Step 6: Calculate the work function: φ = 3.315 x 10^-19 Joules.
Step 7: Convert the energy from Joules to electronvolts (eV) using the conversion factor 1 eV = 1.6 x 10^-19 Joules.
Step 8: Calculate φ in eV: φ ≈ 3.315 x 10^-19 J / (1.6 x 10^-19 J/eV) ≈ 2.07 eV.
Step 9: Round the answer to two decimal places, giving φ ≈ 2.1 eV.

No concepts available.