D. The number of wave that pass a point in a given amount of time
Answer:
λ = 3.1824 10-25 m
Explanation:
To release the electron from the power well the absorbed photon electro must be greater than or equal to the energy of the electron in the power well
The photon energy is E_photon = 0.625 J
The speed of light is
c = λ f
Let's use Planck's equation
E = h f
We substitute
E = h c / λ
λ = h c / E
λ = 6.63 10-34 3 108 / 0.625
λ = 3.1824 10-25 m
Answer:
v = 2 v₀
Explanation:
For this exercise we will use the relationship between work and the change in kinetic energy
W = ΔK
Where the work is of the friction force, which is always opposed to the movement whereby the angle is 180º
W = - fr x
The distance for the first case is
x = 1/4 L
We substitute
-fr ¼ L = ½ m v₀²
fr L = 2 m v₀²
In the second case, the new speed takes the ball the entire distance
x = d
-fr L = ½ m v²
We equal the two equations
2 m v₀² = ½ m v²
v² = 4 v₀²
v = 2 v₀