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ira [324]
3 years ago
6

The potential-energy function u(x) is zero in the interval 0≤x≤l and has the constant value u0 everywhere outside this interval.

an electron is moving past this square well. the electron has energy e=6u0. part a what is the ratio of the de broglie wavelength of the electron in the region x>l to the wavelength for 0
Physics
1 answer:
VMariaS [17]3 years ago
5 0
Look first for the relation between deBroglie wavelength (λ) and kinetic energy (K): 
K = ½mv² 
v = √(2K/m) 
λ = h/(mv) 
= h/(m√(2K/m)) 
= h/√(2Km) 

So λ is proportional to 1/√K. 
in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy: 
K = 6U₀ 

Outer the potential well the potential energy is U₀, so 
K = 5U₀ 
(because kinetic and potential energies add up to 6U₀) 

Therefore, the ratio of the de Broglie wavelength of the electron in the region x>L (outside the well) to the wavelength for 0<x<L (inside the well) is: 
1/√(5U₀) : 1/√(6U₀) 
= √6 : √5
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According to Newton's law of universal gravitation, which statement is true?
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Answer:

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3 years ago
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A charge moves a distance of 1.8 cm in the direction of a uniform electric field having a magnitude of 214 N/C. The electrical p
Andrei [34K]

Answer:

 13.4 x 10 raise to power -19  C

Explanation:

. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m

. The uniform electric field is  E = 214 N/M

, The decrease in electrical potential energy is   d(P.E) = 51.63 x 10 raise to power -19 J

Let the magnitude of the charge of the moving particle be q

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d(P.E) =qEd

51.63 x 10 power -19 = q(214)(0.018)

51.63 x 10 power -19 =3.852q

by making q the formular,

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3 years ago
A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers
Karolina [17]

Answer:

a) 4.9*10^-6

b) 5.71*10^-15

Explanation:

Given

current, I = 3.8*10^-10A

Diameter, D = 2.5mm

n = 8.49*10^28

The equation for current density and speed drift is

J = I/A = (ne) Vd

A = πD²/4

A = π*0.0025²/4

A = π*6.25*10^-6/4

A = 4.9*10^-6

Now,

J = I/A

J = 3.8*10^-10/4.9*10^-6

J = 7.76*10^-5

Electron drift speed is

J = (ne) Vd

Vd = J/(ne)

Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)

Vd = 7.76*10^-5/1.3584*10^10

Vd = 5.71*10^-15

Therefore, the current density and speed drift are 4.9*10^-6

And 5.71*10^-15 respectively

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