All categories

3 years ago

A. Describe the photoelectric effect and explain why it could not be explained by Newtonian physics.

2 answers:

8 0

a) This effect results in emission of electrons out of the lighted area of metals. According to Newtonian mechanics, energy is continuous (may take any value), but photoelectric effect implies using discrete energy ideas.

b) 'Quantum' means 'part' and is being used in physics fir describing light. Light consists of separate particles. This theory fits well what we observe in photoelectric effect: one light particle smashes one electron and it leaves surface of a metal.

c) Quantum mechanics describes everything as wave and particle at a time and it is called dualism. Electrons have separate quantified energetic levels and they go up/down consuming/emitting light waves. The lines in the spectrum fit these energetic levels.

Leno4ka [110]3 years ago

3 0

Explanation:

(a)

The photoelectric effect is the phenomenon in which the light of the particular frequency incidents on the material. Then the emission of the electrons from the surface of the material occurs.

This phenomenon could not be explained by Newtonian physics.

In Newtonian physics, the energy is not discrete. In quantum mechanics, the energy is discrete. This is the main why the photoelectric effect could not be explained by Newtonian physics.

(b)

Light consists of photons. The photon is a packet of energy. It is also called quanta. The energies of the photons are quantized.

When a photon strikes on the surface of metal then the energy of photon is absorbed by an electron in the metal so that it may eject from the surface. This phenomenon is called the photoelectric effect.

(c)

In quantum mechanics, wave-particle duality concept is used to explain the wave-particle nature of the light. Light behaves as particle as well as wave. It shows both nature. The photoelectric phenomenon shows the particle nature of the light. It acts as a particle when it hits the surface of the metal.

In line spectra, the electron is excited to an energy level. In this case energy is transferred from photon to electron. There is a collision between photon and electron. The change in momentum will occur. It shows the particle nature of the light.

You might be interested in

On that system is<br> In an isolated system, the net __<br> zero.

DedPeter [7]

Answer:

the net work is 0

Explanation:

in an isolated system, there is no energy entering or exiting the system, and since work is dependent on the change in energy (W = ΔE) , with no change in energy, there is no work done. the net work done on or by the system must be 0

7 0

2 years ago

A sled is accelerating down a hill at a rate of 1 m s2 . If the mass of the sled is suddenly cut in half and the net force on th

mihalych1998 [28]

We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
$a'= \frac{2F}{ \frac{m}{2} } = \frac{4F}{m} = 4\frac{F}{m}$
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.

4 0

2 years ago

Please help!!! what is the main point of paragraph 3?

Nikitich [7]

There’s no picture:(

8 0

2 years ago

Read 2 more answers

A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and

Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

$q_{out}=q_{2}-q_{1}$

$q_{out}=-6.43-(-4.00)$

$q_{out}=-2.43\ \mu C$

The charge on the inner surface is q.

$q+(-2.43)=-6.43$

$q=-6.43+2.43= 4.00\ \mu C$

We need to calculate the electric field outside the shell

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}$

$E=33927618.73\ N/C$

$E=3.39\times10^{7}\ N/C$

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

5 0

2 years ago

For an atom to be neutral the amount of ? and? must be equal

maks197457 [2]

Answer: Well first of all, for an Atom to be neutral, it shall have an equal number of the protons and the neutrons. So given that, the protons are able to be positively charged and since the electrons are negatively charged (Protons and Electrons are completely different from each other) then that means the Algebric are aggregated of charges because it equal to zero. So overall, the 'neutral atom' has to be neutral, and how? Protons must equal the number of electrons or else it wouldn't be considered neutral.

Hopefully this helps you very much!! :-) Have a great daayy!

6 0

2 years ago

Read 2 more answers