Under what circumstances can energy level transitions occur?

Physics

1 answer:

Yuliya22 [10]3 years ago

8 0

Answer:

Energy transition therefore occurs due to the amount of kinetic energy gained by the electrons. The electrons with higher kinetic energy are excited to the higher level (excited state) compare to the electron with low kinetic energy (this energy are energy in the ground state)

Explanation:

Energy level transition occur when light rays strikes a metal surface to emit electron from the surface, a term known as photoelectric effect. This amount of electron emitted from the surface depends on the speed of light ray striking the metal surface.

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Is baking soda less dense or more dense than water?

boyakko [2]

It’s more dense than air and less dense than liquid!

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3 years ago

Tech A says that a relay is a one-way electrical check valve used in alternators to change AC into DC. Tech B says that a relay

Setler [38]

Answer:

The correct answer is from Tech B.

Explanation:

The relay is an electromagnetic device, which is stimulated by a very weak electrical current to open or close a circuit. It functions as a switch controlled by an electrical circuit in which, by means of a coil and an electromagnet, a sequence of one or several contactors is activated that allow to open or close other electrical circuits independent of the one that controls it.

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3 years ago

A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

$Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}$

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

$E=\dfrac{Q}{4\pi\epsilon_0 r^2}$

Substitute numerical values:

$E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}$

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

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