which research model refers to the study of an individual group or community over a predetermined period
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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
The electric field produced by a single-point charge is given by
where
k is the Coulomb's constant
q is the charge
r is the distance from the charge
To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.
1) The first charge is , and it is located at x=0, so its distance from the point x=0.200 m is
Therefore, the electric field is
And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.
2) The second charge is and it is located at x=0.800 m, so its distance from the point is
Therefore, the electric field is
And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.
3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have
and the sign tells us that the field is directed toward negative x-direction.