Violet light (λ = 430 nm)is incident on a piece of potassium (φ = 2.20 eV).What is the maximum kinetic energy of the ejected photoelectrons? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m)

Question

Quizplus · Accepted Answer

The maximum kinetic energy of the ejected photoelectrons is given by:
KEmax = hν - φ
We know that:
ν = c/λ
Therefore:
ν = (3.00 × 108 m/s)/(430 × 10-9 m)
ν = 6.98 × 1014 Hz
We can substitute this value, along with the given values of h and φ, into the above equation to get:
KEmax = (6.63 × 10-34 J s)(6.98 × 1014 Hz) - (2.20 eV)(1.60 × 10-19 J/eV)
KEmax = 4.62 × 10-19 J - 3.52 × 10-19 J
KEmax = 1.10 × 10-19 J
We can convert this value to electronvolts (eV) by dividing by the conversion factor 1.60 × 10-19 J/eV:
KEmax = (1.10 × 10-19 J)/(1.60 × 10-19 J/eV)
KEmax = 0.69 eV

Therefore, the correct answer is B.

Violet Light (λ = 430 Nm)is Incident on a Piece