Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant
Given:
P = 687 days
A = 1.52 AU
Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³
Answer: 1.3439 x 10⁵ (days²/AU³)
Answer:
Right now I have three.
Explanation: Thanks for the points luv ^-^.
Answer:
The work function ϕ of the metal = 53.4196 x 10⁻¹⁶ J
Explanation:
When light is incident on a photoelectric material like metal, photoelectrons are emitted from the surface of the metal. This process is called photoelectric effect.
The relationship between the maximum kinetic energy (
) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f₀) of the photoemissive metal surface is:
= h(f − f₀)
= hf - hf₀
E is the energy of the absorbed photons: E = hf
ϕ is the work function of the surface: ϕ = hf₀
= E - ϕ
Frequency f = 8.12×10¹⁸ Hz
Maximum kinetic energy
= 4.16×10⁻¹⁷ J
Speed of light c = 3 x 10⁸ m/s
Planck's constant h = 6.63 × 10⁻³⁴ Js
E = hf = 6.63 × 10⁻³⁴ x 8.12×10¹⁸
E = 53.8356 x 10⁻¹⁶ J
from
= E - ϕ ;
ϕ = E - 
ϕ = 53.8356 x 10⁻¹⁶ - 4.16×10⁻¹⁷
ϕ = 53.4196 x 10⁻¹⁶ J
The work function of the metal ϕ = 53.4196 x 10⁻¹⁶ J