Why do atoms absorb and reemit radiation at characteristic frequencies?

1 answer:

7 0

To answer this question it is necessary to use the Bohr Model as a theoretical reference. According to this model, the energy levels of an atom are divided by discrete and characteristic energies. When there is the emission or absorption of a photon with a characteristic energy there is a transition of another energy level, which is equal to the level of atomic energy. Since the energy of a photo is directly proportional to its frequency, the emitted or absorbed photons have characteristic frequencies to the difference in energy between atomic energy levels.

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A +15 nC point charge is placed on the x axis at x = 1.5 m, and a -20 nC charge is placed on the y axis at y = -2.0m. What is th

IceJOKER [234]

Answer: $E=75\ N/m$

Explanation:

Given

First charge of $q_1=15\ nC$ is placed at $x=1.5\ m$

Second charge $q_2=-20\ nC$ is placed at $y=-2\ m$

Electric field is given by

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

Electric field due to $q_1$ is away from it

$E_1=\frac{9\times 10^9\times 15\times 10^{-9}}{(1.5)^2}$

$E_1=60\ N/m$

Electric field due to $q_2$

$E_2=\frac{9\times 10^9\times 20\times 10^{-9}}{2^2}$

$E_2=45\ N/m$

Net electric field will be vector addition of two

$\vec{E_{net}}=\vec{E_1}+\vec{E_2}$

$\vec{E_{net}}=-60\hat{i}-45\hat{j}$

Magnitude of Electric field is

$E=\sqrt{60^2+45^2}$

$E=75\ N/m$

8 0

3 years ago

T/-15=-3. what is t?

sweet-ann [11.9K]

Answer:

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