Answer:
a) 4.15×10∧-15 eV/s
b) 1.31 eV
c) 950.4 nm
Explanation:
let Ф be the work function.
a) f = c/λ = (3×10^8)/(320×10^-9) = 9.375×10^14 /s
f = c/λ = (3×10^8)/(412×10^-9) = 7.28×10^14 /s
E = hf +Ф
Energy which relates to the poteial is calculated in eV (electonvolts).
so E = eV
Now using the straight line analogy in the equation above h is the gradient of the line.
Note that energy which relates to the poteial is calculated in eV (electonvolts).
so E = eV
and h = eV/f
h=[(0.74 - 1.61)eV]/[((7.28 - 9.375)×10^14 )/s]
= 4.15×10^-15 eV/s ≈ 6.65×10^-34 J/s
b) E = hf +Ф
eV = hc/λ - Ф
Ф = hc/λ - eV
= (4.15×10^-15 eV/s)(3×10^8)/[(320×10^-9) - (1.60×10^-19)(1.61)eV]
= 1.31 eV ≈ 2.1×10^-19 J/s
c) Ф = hc/λ0
λ0 = hc/Ф
= [(6.65×10^-34 J/s)(3×10^8 )]/ (2.1×10^-19 J/s)
= 9.50×10^-7 m
