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3 years ago

If we decrease the size of a quantum dot that contains an electron, what happens to the energy levels?

1 answer:

6 0

We will solve this problem using the direct concept related to band gap energy, that is, a band gap is the distance between the valence band of electrons and the conduction band, i. e, the energy range in a solid where no electron states (Electronic state) can exist Mathematically can be described as,

$E_n = \frac{h^2n^2}{8mcR^2}$

Where,

h = Planck's constant

n = Energy level

mc = Effective mass of the point charge

R = Size of the particle

As you can see the energy is inversely proportional to the size of the particle:

$E_n \propto \frac{1}{R^2}$

Therefore if the size is decreased, the amount of energy is increased.

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What is the unit for molar mass?

dmitriy555 [2]

Answer:

kilogram per mole

Explanation:

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3 years ago

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There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target. A pilot fl

nikdorinn [45]

Answer:

Also 3s.

Explanation:

Each component is independent in two dimensional motion. This means that <em>how much time does something take to reach the ground when dropped is independent from any horizontal velocity</em>. If at one run a drop lasts 3s, at another run with twice the (horizontal) velocity and same height will also last 3s, no matter what.

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3 years ago

An igneous rocks color is mainly determined by its

melomori [17]

An ignious rock's color is mainly determined by its silica content

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4 years ago

You read in a science magazine that on the Moon, the speed of a shell leaving the barrel of a modern tank is enough to put the s

olga_2 [115]

To solve this problem we will use the definition of the kinematic equations of centrifugal motion, using the constants of the gravitational acceleration of the moon and the radius of this star.

Centrifugal acceleration is determined by

$a_c = \frac{v^2}{r}$

Where,

v = Velocity

r = Radius

From the given data of the moon we know that gravity there is equivalent to

$a = 1.62m/s$

While the radius of the moon is given by

$r = 1.74*10^6m$

If we rearrange the function to find the speed we will have to

$v = \sqrt{ar}$

$v = \sqrt{1.6(1.74*10^6)}$

$v = 1.7km/s$

The speed for this to happen is 1.7km/s

3 0

3 years ago

A beam of light in air is incident at an angle of 30º to the surface of a rectangular block of clear plastic (n = 1.46). The lig

Aneli [31]

Answer:

θ = 30°

Explanation:

Firts, the angle when the beam of light passes through the block cam be calculated using Snell Law:

$n_{1}sin(\theta_{1}) = n_{2}sin(\theeta_{2})$

<u>Where</u>:

n₁: is the index of refraction of the incident medium (air) = 1

θ₁: is the incident angle = 30°

n₂: is the medium 2 (plastic) = 1.46

θ₂: is the transmission angle

Hence, θ₂ is:

$sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1*sin(30)}{1.46} = 0.34 \rightarrow \theta_{2} = 20.03 ^{\circ}$

Now, when the beam of light re-emerges from the opposite side, we have:

n₁: is the index of refraction of the incident medium (plastic) = 1.46

θ₁: is the incident angle = 20.03°

n₂: is the medium 2 (air) = 1

θ₂: is the transmission angle

Hence, the angle to the normal to that surface (θ₂) is:

$sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1.46*sin(20.03)}{1} = 0.50 \rightarrow \theta_{2} = 30 ^{\circ}$

Therefore, we have that the beam of light will come out at the same angle of when it went in, since, it goes from air and enters to a plastic medium and then enters again in this medium to go out to air again. This was proved using the Snell Law.

I hope it helps you!

5 0

3 years ago