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aliina [53]
3 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]3 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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1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position
antiseptic1488 [7]

Answer:

The value of the distance is \bf{14.52~cm}.

Explanation:

The velocity of a particle(v) executing SHM is

v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)

where, \omega is the angular frequency, A is the amplitude of the oscillation and x is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., x = 0.

The maximum velocity(\bf{v_{m}}) is

v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Divide equation (1) by equation(2).

\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)

Given, v = 0.25 v_{m} and A = 15~cm. Substitute these values in equation (3).

&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm

6 0
3 years ago
How can you use graphs to calculate the displacement of an object?
tekilochka [14]

Answer:

c. find the slope of the velocity time graph

8 0
2 years ago
I need help as fast as possible please ITs DUE IN 10 MINUTES
algol [13]

Answer:

the image is behind the mirror

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3 0
2 years ago
The energy of random atomic and molecular motion is called.
Vanyuwa [196]
Energy of a random atomic and molecular is called molecules energy
5 0
2 years ago
In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de
snow_tiger [21]

As we know that range of the projectile motion is given by

R = \frac{v^2 sin(2\theta)}{g}

here we know that range will be same for two different angles

so here we can say the two angle must be complementary angles

so the two angles must be

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so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

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2 years ago
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