Answer:a) λ = 4.862 10⁻⁷ m, b) λ = 4.341 10⁻⁷ m
Explanation:
The spectrum of hydrogen can be described by the expression
in the case of the initial state n = 2 this series is the Balmer series
a) Find the wavelength for n = 4
let's calculate
= 1,097 10⁷ ()
\frac{1}{ \lambda} = 1.097 10⁷ 0.1875 = 0.2056 10⁷
λ = 4.862 10⁻⁷ m
b) n = 5
\frac{1}{ \lambda} = 1,097 10⁷ ()
\frac{1}{ \lambda} = 1.097 10⁷ 0.21 = 0.23037 10⁷
λ = 4.341 10⁻⁷ m
Im pretty sure that the first modle had the atom fully solid with no gaps!
Answer:
E = 5.03 10⁻³ N / C
Explanation:
For this exercise we can use Gauss's law
Ф = E. dA = /ε₀
Where as a Gaussian surface we select a cylinder
Ф= E A = q_{int} /ε₀
Let's use the concept of surface charge density
σ = q / A
q = σ A
E A = σ A /ε₀
E = sig /ε₀
Since the plate is non-conductive, the fux does not flow through the plate
With this equation the electric field does not depend on the distance, so the field at 1 cm
E = 45 10³ N/C
σ = E ε₀
σ = 45 10³ 8.85 10⁻¹²
σ = 3.98 10⁻¹⁰ C / m²
As the point where the field is requested is very far from the plate we can not ignore the finite size of it
At r = 15 m the plate has a very small size, so let's calculate the field as created by a point load at this distance
E = k q / r²
The plate area is
A = L L
A = 0.75²
A = 0.5625 m²
q = σ A
q = 3.98 10⁻¹⁰ 0.5625
q = 1.26 10⁻¹⁰ C
Let's calculate
E = 8.99 10⁹ 1.26 10⁻¹⁰ / 15²
E = 5.03 10⁻³ N / C
the complete question in the attached figure
we have that
d=0.51 mm------------------------- >0.00051 m
v = √(2ad) = √(2 * 1300m/s² * 0.00051 m) = 1.15 m/s initial velocity
a=9.8 m/s²
Then d = v² / 2a = (1.15m/s)² / (2*9.8) m/s² = 0.059 m = 59 mm
the answer is 59 mm