Answer:
a) E₀ = 2.125 eV, b) # photon2 = 9.2 10¹⁵ photons / mm²
Explanation:
a) To calculate the energy of a photon we use Planck's education
E = h f
And the ratio of the speed of light
c = λ f
We replace
E = h c /λ
Let's calculate
E₀ = 6.63 10⁻³⁴ 3 10⁸/585 10⁻⁹
E₀ = 3.40 10⁻¹⁹ J
Let's reduce
E₀ = 3.4 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
E₀ = 2.125 eV
b) Let's look for the energy in each pulse
P = E / t
E = P t
E = 20.0 0.45 10⁻³
E = 9 10⁻³ J
let's use a ratio of proportions (rule of three) if we have the energy of a photon (E₀), to have the energy of 9 10⁻³ J
# photon = 9 10⁻³ /3.40 10⁻¹⁹
# photon = 2.65 10¹⁶ photons
Let's calculate the areas
Focus area
A₁ = π r²
A₁ = π (3.4/2)²
A₁ = 9,079 mm²2
Area requested for calculation r = 1 mm
A₂ = π 1²
A₂ = 3.1459 mm²
Let's use another rule of three. If we have 2.65 106 photons in an area A1 how many photons in an area A2
# photon2 = 2.65 10¹⁶ 3.1459 / 9.079
# photon2 = 9.2 10¹⁵ photons / mm²
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In short, Your Answer would be Option B
Hope this helps!
