All categories

2 years ago

Prob. 3: Manifestation of quantum phenomena (total 25 points) (a)-(e) 5 points each.

Physics

1 answer:

TEA [102]2 years ago

6 0

Answer:

3A. This phenomenon can be seen in the discrete emission of the molecules.

3B. The emotion of the atoms is observed, from states high in energy to a state of minimum energy that is stable indefinitely.

3C. When an electron beam passes through an inhomogeneous magnetic field, it is divided into only two beams

3D. This is due to the stimulated emission

3E. The penetration of a potential barrier is observed in the radioactive emission of heavy atoms, where an alpha particle (Helium nucleus)

Explanation:

This problem asks for some experimental explanations of various quantum phenomena.

3A. This phenomenon can be seen in the discrete emission of the molecules.

In the classical explanation all states or energies are allowed, therefore when emitting energy (photons) there should be a continuum, this is not observed

In the correct quantum explanation only some states are allowed, therefore the emission must be discrete, which is observed in the emission or absorption of molecules and atoms

3B. The emotion of the atoms is observed, from states high in energy to a state of minimum energy that is stable indefinitely.

The incorrect classical explanation that if the minimum energy was zero the electrons cannot rotate around the nuclei and the atom collapses, this does not happen

3C. When an electron beam passes through an inhomogeneous magnetic field, it is divided into only two beams, which is evidence of the existence of two discrete states that we call spin, remember that a free electron beam has zero angular momentum.

3D. This is due to the stimulated emission that occurs when a photon passes through the emission zone, causing the atoms to have transitions and these emitted photons have the same initial photon location, the laser beam all photons are in phase and therefore it is coherent .

This is widely used for holographic and interference work

3E. The penetration of a potential barrier is observed in the radioactive emission of heavy atoms, where an alpha particle (Helium nucleus) leaves the atomic nucleus penetrating the barrier since its energy is lower than the nuclear barrier potential.

You might be interested in

Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

$q_1 = 21.0\mu C$

$q_1 = 47.0\mu C$

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

$E = \frac{kq}{x}^2$

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

$\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2$

solving for x we get

$\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}$

x = 0.200 m

b)electric field at half way mean x =0.25

$E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}$

E = 3.84*10^{-4} N/C

6 0

3 years ago

Read 2 more answers

A 300 N force acts on a 25 kg object. The acceleration of the object is?

storchak [24]

Acceleration is F/M so the answer would be 12m/s^2

7 0

3 years ago

Read 2 more answers

Match the term with each description:

Alex_Xolod [135]

Answer:

An acid is a compound that increases hydrogen ions (H+) when it is dissolved in a solution.

all I got

4 0

3 years ago

What is the name of the force that opposes, or the opposite of gravity ???!!

Novosadov [1.4K]

Answer:

Tension.

<em><u>tension</u></em> is the name of force that opposes or goes opposite of gravity

Hope this helps!

4 0

3 years ago

Read 2 more answers

a ball is thrown straight up into the air with a speed of 13 m/s. if the ball has a mass of 0.25 kg, how high does the ball go?

evablogger [386]

<h2>Hello!</h2>

The answer is: 8.62m

There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.

<h2>First moment:</h2>

Before the ball is thrown, where the potential energy is 0.

<h2>Second moment: </h2>

After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.

Therefore,

$E=PE+KE$

Where:

$PE=m.g.h$

$KE=\frac{1*m*v^{2}}{2}$

<em>E</em> is the total energy

<em>PE</em> is the potential energy

<em>KE</em> is the kinetic energy

<em>m</em> is the mass of the object

<em>g</em> is the gravitational acceleration

<em>h </em>is the reached height of the object

<em>v</em> is the velocity of the object

Since the total energy is always constant, according to the Law of Conservation of Energy, we can write the following equation:

$KE_{1}+PE_{1}=KE_{2}+PE_{2}$

Remember, at the first moment the PE is equal to 0 since there is not height, and at the second moment, the KE is equal to 0 since the velocity at maximum height is 0.

$\frac{1*m*v^{2}}{2}+m.g.(0)=\frac{1*m*0^{2}}{2}+m.g.h\\\frac{1*m*v_{1} ^{2}}{2}=m*g*h_{2}$

So,

$h_{2}=\frac{1*m*v_{1} ^{2}}{2*m*g}\\h_{2}=\frac{1*v_{1} ^{2}}{2g}=\frac{(\frac{13m}{s})^{2} }{2*\frac{9.8m}{s^{2}}}\\h_{2}=8.62m}$

Hence,

The height at the second moment (maximum height) is 8.62m

Have a nice day!

5 0

3 years ago