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marysya [2.9K]
3 years ago
12

What happens when an electron moves from an excited state to the ground state?

Physics
1 answer:
Darina [25.2K]3 years ago
6 0
<span>When an electron moves from an excited state to the ground state, "Energy releases"

Hope this helps!</span>
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A concise definition of pair production
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Pair production<span> is a direct conversion of radiant energy to matter. It is one of the principal ways in which high-energy gamma rays are absorbed in matter. </span>
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krek1111 [17]

Explanation:

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A person standing a certain distance from four identical loudspeakers is hearing a sound level intensity of 125 dB. What sound l
DENIUS [597]

Answer:

\mathbf{\beta = 123.75 \ dB}

Explanation:

From the question, using the expression:

125 \ dB = 10 \ log (\dfrac{I}{I_o})

where;

I_o = 10^{-12} \ W/m^2

I = 10^{12.5} \times 10^{-12} \ W/m^2

I = 3.162 \ W/m^2

This is a combined intensity of 4 speakers.

Thus, the intensity of 3 speakers = \dfrac{3.162\times 3}{4}

= 2.372 W/m²

Thus;

\beta = 10 \  log ( \dfrac{2.372}{10^{-12}} ) \ W/m^2

\mathbf{\beta = 123.75 \ dB}

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3 years ago
Circular motion formulas
KIM [24]

Answer:

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5 0
3 years ago
A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.
kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

6 0
3 years ago
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