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ira [324]
4 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]4 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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5) A 20.0 kg cart with no friction wheels sits on a table. A light string is attached to it and runs over a low friction pulley
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Answer:

1) Please find attached, created with Microsoft Visio

2) The acceleration of the masses connected by the light string is 0.00735 m/s²

3) The tension in the cord is 0.147 N

4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s

5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s

Explanation:

1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio

2) The acceleration of the masses connected by the light string is given as follows;

F = Mass, m × Acceleration, a

The mass of the truck, M = 20.0 kg

The mass attached to the string, hanging rom the pulley, m = 0.0150 kg

The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N

The pulling force acting on the cart, F = M × a

∴ F = 0.147 N = 20.0 kg × a

a = 0.147 N/(20.0 kg) = 0.00735 m/s²

The acceleration of the truck = a = 0.00735 m/s²

3) The tension in the cord = F = 0.147 N

4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²

Where;

s = The distance to the edge of the table = 1.2 m

u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)

a = The acceleration of the cart = 0.00735 m/s²

t = The time taken

Substituting the known values, gives;

s = u·t + 1/2·a·t²

1.2 = 0 × t + 1/2 ×0.00735 × t²

1.2 = 1/2 ×0.00735 × t²

t² = 1.2/(1/2 ×0.00735) ≈ 326.5306

t = √(1.2/(1/2 ×0.00735)) ≈ 18.07

The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s

5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;

v² = u² + 2·a·s

u = 0 m/s

v² = 0² + 2 × 0.00735 × 1.2 = 0.001764

v = √(0.001764) = 0.042

The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.

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