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
Air Pressure is a force that is exerted onto your body by air molecules.
Bohr's equation for the change in energy is

where
h = Planck's constant
c == the velocity of light
λ = wavelength.
The velocity is related to wavelength and frequency, f, by
c = fλ
Let us examine the given answers on the basis of the given equations.
a. As λ increases, f decreases and ΔE decreases.
TRUE
b. As λ increases, f increases and ΔE increases.
FALSE
c. As λ increases, f increases and ΔE decreases.
FALSE
Answer:
As the wavelength increases, the frequency decreases and energy decreases.
Answer:
Time interval;Δt ≈ 37 seconds
Explanation:
We are given;
Angular deceleration;α = -1.6 rad/s²
Initial angular velocity;ω_i = 59 rad/s
Final angular velocity;ω_f = 0 rad/s
Now, the formula to calculate the acceleration would be gotten from;
α = Change in angular velocity/time interval
Thus; α = Δω/Δt = (ω_f - ω_i)/Δt
So, α = (ω_f - ω_i)/Δt
Making Δt the subject, we have;
Δt = (ω_f - ω_i)/α
Plugging in the relevant values to obtain;
Δt = (0 - 59)/(-1.6)
Δt = -59/-1.6
Δt = 36.875 seconds ≈ 37 seconds