If Mars were the same size as Mercury (instead of its actual size), which surface features would it have?
1 answer:
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Answer:
When both initial speed and initial displacement is doubled then amplitude will be doubled.
Explanation:
Given that :- Amplitude of simple harmonic Oscillator is doubled.
So,
Formula of Simple harmonic oscillator is ...........(1)
Where X = Position in (m,cm,km.....)
A = Amplitude in (m,cm,km.....)
F = Frequency in (Hz)
T = Time in (sec.)
Ф = Phase in (rad)
For initial displacement taking t=0 we get,
Initial displacement = .................(2)
Now taking equation (1) and differentiating it w.r.t to (t) we get
taking t=0 for initial speed then we get,
Initial speed = ...............(3)
observing equation (2) & (3) that the initial displacement and initial speed depends on the Amplitude of the Oscillator.
Hence,
when both initial speed and displacement is doubled then amplitude will be doubled.
Answer:
tex]\lambda_{Be}[/tex] = 22.78 nm
Explanation:
Bohr's model for the hydrogen atom has been used by other atoms with a single electric charge by changing the number of charges by the charge of the new atom (atomic number)
= k e² / 2a₀ (1 /n²)
ao = h'² / k m e² h' = h/2πi
For another atom with a single electron in the last layer
a₀ ’= h’² / k m (Ze)²
a₀ ’= a₀ / Z²
Therefore, when replacing in the equation
= - Z² Eo/n²
E₀ = 13,606 eV
The transition occurs when the electron stops from one level to another
-
= Z² E₀ (1 / n² - 1 / m²) = Z² ΔE
Let's relate this expression to the wavelength
c = λ f
E = h f
E = h c /λ
h c / λ = Z² ΔE
λ = 1 / Z² (hc / ΔE)
λ = 1 / Z² λ_hydrogen
Let's apply this last equation to our case
Lithium Z = 3
= - 9 Eo / n²
40.5 10-9 = 1/9 λ_hydrogen
Beryllium Z = 4
λ = 1/16 λ_hydrogen
Let's write our two equations is and solve
40.5 10-9 = 1/9 λ_hydrogen
tex]\lambda_{Be}[/tex] = 1/ 16 λ_hydrogen
40.5 10⁻⁹ = 1/9 (16 )
tex]\lambda_{Be}[/tex] = 40.5 9/16
tex]\lambda_{Be}[/tex] = 22.78 nm