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sergey [27]
3 years ago
12

When is the absorption spectrum of an electron generated

Physics
2 answers:
serious [3.7K]3 years ago
7 0
A
Excitation to a higher energy state requires energy which is absorbed from the electromagnetic waves applied.
Goshia [24]3 years ago
6 0

Answer: A. when an electron in an atom is excited to a higher energy state

Explanation:

An absorption spectrum consists of dark lines on a visible background. The dark lines indicate the wavelength which is absorbed by the electron to jump from lower energy state to higher energy state.

Electron requires energy to jump to higher state. Specific wavelength is absorbed by the electron which appears as dark line. This wavelength is equivalent to energy difference between the two levels the electron is jumping.

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It is called the reaction force of a bird flying.
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The container is filled to yh 350 ml mark water is observation or inference?
Lorico [155]
This is an observation.
4 0
3 years ago
You want to produce three 2.00-mm-diametercylindrical wires, each with a resistance of 1.00 Ω at room temperature. One wire is g
Vlada [557]

Answer:

(a) L =  128.75 m

(b) L =  182.56 m

(c) L =  114.28 m

(d) Mass of Gold = 7.68 kg = 7680 gram

(e) Cost of Gold Wire = $ 307040

Explanation:

The resistance of the wire is given as:

R = ρL/A

where,

R = Resistance

ρ = resistivity

L = Length

A = cross-sectional area

(a)

For Gold Wire:

ρ = 2.44 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)

<u>L =  128.75 m</u>

<u></u>

(b)

For Copper Wire:

ρ = 1.72 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)

<u>L =  182.56 m</u>

<u></u>

(c)

For Aluminum Wire:

ρ = 2.75 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)

<u>L =  114.28 m</u>

<u></u>

(d)

Density = Mass/Volume

Mass = (Density)(Volume)

Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³

Therefore,

Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)

<u>Mass of Gold = 7.68 kg = 7680 gram</u>

<u></u>

(e)

Cost of Gold Wire = (Unit Price of Gold)(Mass of Gold)

Cost of Gold Wire = ($ 40/gram)(7680 grams)

<u>Cost of Gold Wire = $ 307040</u>

7 0
4 years ago
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