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andrey2020 [161]
3 years ago
13

A material kept at high temperature is seen to emit photons with energies of 0.3 eV, 0.5 eV, 0.8 eV, 2.0 eV, 2.5 eV, and 2.8 eV.

These are the only photon energies observed. It is now cooled down to a very low temperature so that it is not emitting photons anymore. If a beam of light with a continuous range of energies from 0.01 eV to 10 eV shines on the material, what photon energies in this beam will correspond to dark absorption lines?
Physics
1 answer:
sukhopar [10]3 years ago
5 0

Answer:

0.3 eV, 0.5eV,, 8 eV, 2.0eV, 2.50 eV, 2.8 eV

Explanation:

In a given material the emission and absorption spectra are equivalent, for which the emission spectrum observed at high temperature for the material corresponds to the transition between the energy states of the material, the process is that the electrons exist from the ground state until an excited state and after a short period of time or these electrons relax emitting photons.

In the absorption process, the material is at low temperature, ideally at A = 0K, whereby all states are in the ground state and all excited states are empty. therefore it can absorb the beam energy for each transition given from the ground state to each excited edtado.

Consequently, the lines above the absorption oscillate lines coincide with the lines of emotion, this we see lines oscillate at 0.3 eV, 0.5eV,, 8 eV, 2.0eV, 2.50 eV, 2.8 eV

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Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

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I just took this test, and can 100% confirm this is the proper answer.

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A tennis player smashes a ball of mass m horizontally at a vertical wall. The ball rebounds at the same speed v with which it st
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Answer:

<em> B.0</em>

Explanation:

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Change in momentum = M(V-U)......................... Equation 1

where M = mass of the ball, V = final velocity of the ball, U = initial velocity of the ball.

Let: M = m kg and V = U = v m/s

Substituting these values into equation 1

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3 years ago
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kipiarov [429]

Answer:

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Explanation:

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The density of the woman is calculate as;

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Density of the woman = 950.8 kg/m³

Therefore, the density of the woman is 950.8 kg/m³

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