A double-slit experiment uses light of wavelength 650 nm with a slit separation of 0.100 mm and a screen placed 4.0 m away. a) W
1 answer:
Answer:
Explanation:
a ) Slit separation d = .1 x 10⁻³ m
Screen distance D = 4 m
wave length of light λ = 650 x 10⁻⁹ m
Width of central fringe = λ D / d
=
= 26 mm
b ) Distance between 1 st and 2 nd bright fringe will be equal to width of dark fringe which will also be equal to 26 mm
c ) Angular separation between the central maximum and 1 st order maximum will be equal to angular width of fringe which is equal to
λ / d
=
= 6.5 x 10⁻³ radian.
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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