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aliina [53]
2 years ago
10

How do the dark lines of an atom''s absorption spectrum relate to the bright lines of its emission spectrum?

Physics
1 answer:
tangare [24]2 years ago
3 0

Wouldn't it be neat if an electron falling closer to the nucleus ... emitting a
photon ... actually gave out more energy than it needed to climb to its original
energy level by absorbing a photon !   If there were some miraculous substance
that could do that, we'd have it made.

All we'd need is a pile of it in our basement, with a bright light bulb over the pile,
connected to a tiny hand-crank generator.

Whenever we wanted some energy, like for cooking or heating the house, we'd
switch the light bulb on, point it towards the pile, and give the little generator a
little shove.  It wouldn't take much to git 'er going.

The atoms in the pile would absorb some photons, raising their electrons to higher
energy levels.  Then the electrons would fall back down to lower energy levels,
releasing more energy than they needed to climb up.  We could take that energy,
use some of it to keep the light bulb shining on the pile, and use the extra to heat
the house or run the dishwasher.

The energy an electron absorbs when it climbs to a higher energy level (forming
the atom's absorption spectrum) is precisely identical to the energy it emits when
it falls back to its original level (creating the atom's emission spectrum).

Energy that wasn't either there in the atom to begin with or else pumped
into it from somewhere can't be created there.

You get what you pay for, or, as my grandfather used to say, "For nothing
you get nothing."

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What is The substance that dissolves the solute.
Ludmilka [50]

Answer:

Explanation:

Do you mean the solvent? If this is off the mark, let me know in a comment.

The solvent is something that the solute is (usually) soluble in.

3 0
2 years ago
In a parallel portion of a series-parallel circuit, the voltage across the branches can be found by multiplying the sum of the b
Julli [10]

Answer:

It's true.

Explanation:

It's true. When we connect two resistors in parallel the current is divided between the two in such a way that the sum of the currents on each resistor should be equal to the current on that branch. By finding the equivalent resistance we can use Ohm's law to determine the voltage drop across the resistors. This voltage drop is the same for both, since they're connected in parallel.

8 0
3 years ago
When two objects collide, the momentum of the system remains constant. This is a statement of ______________.
Fudgin [204]
Conservation of momentum
3 0
3 years ago
Read 2 more answers
A noninverting op-amp circuit with a gain of 96 V/V is found to have a 3-dB frequency of 8 kHz. For a particular system applicat
umka21 [38]

Answer:

Av_2 =24\ V/V

Explanation:

given,

op-amp circuit with a gain of = (Av₁) = 96 V/V

Band width  = (Bw₁) = 8 kHz

Required bandwidth(Bw₂) = 32 kHz

Highest gain available =(Av₂) = ?

For the given system Bandwidth product is constant

                           Av₁ Bw₁ = Av₂ Bw₂

                           96 x 8 = Av₂ x 32

                           Av_2= \dfrac{96\times 8}{32}

                           Av_2 =24\ V/V

the highest gain available under these conditions Av_2 =24\ V/V

6 0
3 years ago
A wave traveling in the positive x-direction with a frequency of 50.0 Hz is shown in the figure below. Find the following values
Klio2033 [76]

Answer:

Explanation:

a. The amplitude is the measure of the height of the wave from the midline to the top of the wave or the midline to the bottom of the wave (called crests). The midline then divides the whole height in half. Thus, the amplitude of this wave is 9.0 cm.

b. Wavelength is measured from the highest point of one wave to the highest point of the next wave (or from the lowest point of one wave to the lowest point of the next wave, since they are the same). The wavelength of this wave then is 20.0 cm. or \lambda=20.0cm

c. The period, or T, of a wave is found in the equation

f=\frac{1}{T} were f is the frequency of the wave. We were given the frequency, so we plug that in and solve for T:

50.0=\frac{1}{T} so

T=\frac{1}{50.0} and

T = .0200 seconds to the correct number of sig fig's (50.0 has 3 sig fig's in it)

d. The speed of the wave is found in the equation

f=\frac{v}{\lambda} and since we already have the frequency and we solved for the wavelength already, filling in:

50.0=\frac{v}{20.0} and

v = 50.0(20.0) so

v = 1.00 × 10³ m/s

And there you go!

5 0
2 years ago
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