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kipiarov [429]
3 years ago
9

What has research determined about the orbit of an electron around a nucleus?

Physics
2 answers:
sergeinik [125]3 years ago
8 0
The one that research has determined about the orbit of an electron around nucleus is : Each sub-level electron type has a unique path where it will likely to be found
Here are the sub levels of an electron :
-sub level s, maximum number of 2 electrons
- sub level p, maximum number of 6 electrons
- sub level d, maximum number of 10 electrons
- sub level f, maximum number of 14 electrons
melamori03 [73]3 years ago
4 0

Answer:

Each sub-level electron type has a unique path where it will likely be found.

Explanation:

The quantum atomic model is the actual model of an atom. Electrons are present in certain regions called as orbitals. Each orbitals corresponds to some sub levels of energy. These are identified with letters s,p,d and f.

Each orbitals have different shapes like s orbitals are spherical, p orbitals have dumbbell shape etc.

The orbit of an electron around a nucleus is like unique path where it will likely to found. So, the correct option is (c).

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Which statement(s) correctly compare the masses of protons, neutrons, and electrons?
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Electron<span>. the central part of an atom containing </span>protons<span> and </span>neutrons<span> ... which of the following is necessary to calculate the atomic </span>mass<span> of an element? ... which of the </span>statements correctly compares<span>the relative size of an ion to its neutral atom?</span>
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How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
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Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0
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Answer:

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