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Ann [662]
3 years ago
11

The ground state energy of an electron in a one-dimensional trap with zero potential energy in the interior and infinite potenti

al energy at the walls is 2.0 eV. If the width of the well is doubled, the ground state energy will be:
Physics
1 answer:
ElenaW [278]3 years ago
6 0

Answer:

0.5 eV

Explanation:

E_1 = Initial potential energy = 2\ eV

E_2 = Final potential energy

L_1 = Initial width

L_2 = Final width = 2L_1

Energy of an electron in a one-dimensional trap is given by

E=\dfrac{n^2h^2}{8mL^2}

From the equation we get

E\propto \dfrac{1}{L^2}

So,

\dfrac{E_1}{E_2}=\dfrac{L_2^2}{L_1^2}\\\Rightarrow E_2=\dfrac{E_1L_1^2}{L_2^2}\\\Rightarrow E_2=\dfrac{2L_1^2}{4L_1^2}\\\Rightarrow E_2=0.5\ eV

The ground state energy will be 0.5 eV

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3. A football is kicked with a speed of 35 m/s at an angle of 40°.
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a) 22.5 m/s

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u_y = u sin \theta

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u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

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u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

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where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

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3 years ago
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