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zvonat [6]
3 years ago
5

When the atom's electrons step down to lower energy levels in a thin cloud of hot gas, what is produced?

Physics
2 answers:
Reika [66]3 years ago
8 0

Answer: The correct answer is an emission line spectrum.

Explanation:

When the electrons are excited to the higher energy level, the energy is absorbed in this case.

When the electrons in the atom step down to lower energy levels in a thin cloud of hot gas then the radiation will emit.

The electron will lose energy in this case in the form of radiation. There will be an emission line spectrum.

ryzh [129]3 years ago
7 0

When the electrons step down to lower energy level, emission line spectrum produced.

Further Explanation:

The spectrum of emission of an element is spectrum of wavelength of radiation emitted during the transition of electron from a higher energy level to lower energy level. The energy of photon emitted in this process is equal to difference between the two levels. There is possibility of different types of transition for an atom and energy difference is specific for each atom.

When electrons are excited in atom, extra energy forces electrons to higher energy orbital. When electron goes to lower state, energy releases in form of photon. This emitted photon form the emission line spectrum.

Emission is process in which higher energy state of the particle convert into a lower state through the emission of photon, which produces light. The frequency of emitted light is function of transitional energy.

Since energy of system conserved, difference in energy between two states equal to energy carried by photon. The energy state of transitions results in emission of large frequency range.

Learn more:

1. Broadcast wavelength of station: brainly.com/question/9527365

2. Stress under the tension: brainly.com/question/12985068

3. Average translational kinetic energy: brainly.com/question/9078768

Answer Details:

Grade: High School

Subject: Physics

Chapter: Modern physics

Keywords:

Electrons, step, lower, energy, levels, thin, cloud, hot, gas, emission, line, spectrum, ground, state, higher, radiation, transition, energy, light and photon.

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Based on the following equation, answer the questions below. ρ = (2γϕ + ψ)/rg where ρ [=] moles per cubic foot [mol/ft3] γ [=] j
AlekseyPX

1) Fundamental units of \Psi are [\frac{mol}{m\cdot s^2}]

2) Fundamental units of \Phi are [\frac{mol}{m^3}]

Explanation:

The equation for the variable \rho is

\rho =\frac{2\gamma \Phi+\Psi}{rg}

where we have:

\rho measured in [\frac{mol}{ft^3}]

\gamma measured in [\frac{J}{kg}]

r measured in [in]

g measured in [\frac{m}{s^2}]

We can re-write the equation as

\rho rg = 2\gamma \Phi + \Psi

And we notice that the units of the term on the left must be equal to the units of the term on the right.

This means that:

1) First of all, \Psi must have the same units of \rho r g. So,

[\rho r g]=[\frac{mol}{ft^3}][in][\frac{m}{s^2}]

However, both ft (feet) and in (inches) are not fundamental dimensions: this means that they can be expressed as meters. Therefore, the fundamental units of \Psi are

[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]

2)

The term 2\gamma \Phi must have the same units of \Psi in order to be added to it. Therefore,

[\gamma \Phi] = [\frac{mol}{m\cdot s^2}]

We also know that the units of \gamma are [\frac{J}{kg}], therefore

[\frac{J}{kg}][\Phi]= [\frac{mol}{m\cdot s^2}]

And so, the fundamental units of \Phi are

[\Phi]= [\frac{mol\cdot kg}{J\cdot m\cdot s^2}]

However, the Joules can be written as

[J]=[kg][\frac{m^2}{s^2}]

Therefore

[\Phi]= [\frac{mol\cdot kg}{(kg \frac{m^2}{s^2})\cdot m\cdot s^2}]=[\Phi]= [\frac{mol}{m^3}]

#LearnwithBrainly

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3 years ago
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