Ask question

Ask question

All categories

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:

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

You might be interested in

Pls help! Fill in the blanks.

mart [117]

Answer:

attracting iron and producing a magnetic field

5 0

2 years ago

An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by p

nexus9112 [7]

Complete Question

An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed?

A It should stay the same

B It should be quadrupled.

C It should be quintupled

D It should be doubled.

E It should be tripled

Answer:

Option D is the correct option

Explanation:

Generally electric field is mathematically represented as

$E = \frac{\sigma}{\epsilon_o}$

Where $\sigma$ is the charge per unit area (Charge density )

From the question we are told that $\sigma$ is doubled hence the

$E = \frac{2 \sigma }{\epsilon_o}$

Looking the equation above we see that the value of the electric field will also double given that it is directly proportional to the charge density

8 0

2 years ago

A ball thrown vertically upward reaches a certain height and comes down again. What can you say about its kinetic energy at the

MrRissso [65]

The K.E. at the maximum height is zero, no matter what that height is.

4 0

3 years ago

Read 2 more answers

A machine has an efficiency of 20%. Find the input work if the output work is 140J.

Ilia_Sergeevich [38]

Ummmm I don’t know this sorry

3 0

2 years ago

A child's water pistol shoots water through a 1.0-mm-diameter hole. If the pistol is fired horizontally 70 cm above the ground,

lesya692 [45]

Answer:

2.52 ml/s

Explanation:

Unit conversions:

1 mm = 0.001m

70 cm = 0.7 m

Let g = 10m/s2. If the pistol is fired horizontally at first, it did not have an vertical velocity, only horizontal velocity. So g is the only thing that affects the vertical motion of water.

We can calculate the time it takes for the squirts to hit the ground in the following equation of motion:

$h = gt^2/2$

where h = 0.7 m is the vertical distance, and t is the time it takes, which is what we are solving for:

$t^2 = 2h/g = 2*0.7/10 = 0.14$

$t = \sqrt{0.14} = 0.374 s$

So the squirts takes 0.374s to hit the ground, and within that time it travels a distance of 1.m horizontally. Neglect air resistance, we can calculate the horizontal velocity:

$v = s/t$

where s = 1.2 m is the horizontal distance

$v = 1.2/0.374 = 3.21 m/s$

The cross-section area of the hole is

$A = \pi r^2$

where r = d/2 = 0.001/2 = 0.0005 m is the radius of the hole

$A = 0.0005^2 \pi = 7.85*10^{-7}$

So we can calculate the volume flow rate:

$\dot{V} = v*A = 3.21*7.85*10^{-7} = 2.52 * 10^{-06} m^3/s$

or 2.52 ml/s

8 0

3 years ago