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
When you work out it helps clear your thoughts and think of nothing. It makes you feel good.
Answer:
Frequency = 260 Hz and wavelength = 1.31 m
Explanation:
Given that,
A hammer begins vibrating back and forth at approximately 260 cycles per second.
(a) The frequency of an object is the number of vibrations per unit time. The frequency of the sound wave is 260 Hz.
(b) The speed of sound in air is 343 m/s. So,
Hence, this is the required solution.
Answer:
No.
Explanation:
Given that Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change ?
Considering the formula
V = wr
Where
V = linear speed
W = angular speed
r = radius of the wheel.
But W = 2πrf
Where the the 2 and pi are constant. The radius of the first wheel will be small but counter balance with the larger frequency.
While the radius of the second wheel may be large but it will be of a small frequency.
We can therefore conclude that the reading on the speedometer will not change. Because speedometer will read the linear speed V.
Since they are both examples of moving waves, they both transmit energy.
