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zmey [24]
3 years ago
9

Consider the following neutral electron configurations in which n has a constant value. Which configuration would belong to the

element with the most negative electron affinity, E ea E_ea ?
2s2
2s2 2p2
2s2 2p5
2s2 2p6
Physics
2 answers:
kramer3 years ago
4 0

Answer:

2s2 2p6.

Explanation:

Electron affinity is a term used to measure of the attraction between the incoming electron and the nucleus - the stronger the attraction, the more energy is released. The factors which affect this attraction are: nuclear charge, distance, electronic configuration and screening.

Comparing the following electronic configuration of their outermost shell:

2s2

2s2 2p2

2s2 2p5

2s2 2p6

Generally, Eae increases across the period(more positive), where the noble gases are the most negative. This is because of their fully filled shells.

Therefore, 2s2 2p6 has fully filled orbitals, therefore it will have the highest negative electron affinity.

grandymaker [24]3 years ago
3 0

Answer:

he configuration with the highest electronic affinity is 2s2 2p5

Explanation:

Electronic affinity is the variation of energy when we add an electron to a neutral atom to form an ion

When an electron is added, it must occupy a space is the sub-level of the atom, giving more stability when it approaches the configuration of a complete shell with eight electrons (noble gas), so the affinity must increase when moving in a period Group VIII noble gases)

Let's examine the given settings

In this case, when adding an electron, 2s2 is very far from a complete level configuration, so its affinity must be small.

2s2 2p2 when adding an electro the one has a little more affinity, but is still a long way from a full shell, it would be missing 3 electrons

2s2 2sp5 this is the atom with the highest electronic affinity, since i = that when adding an electron the ion has the configuration of a noble gas. This is the most stable on the list

2s2 2p6 already has a full shell making it very difficult to insert an electron into this atom.

In summary, the configuration with the highest electronic affinity is 2s2 2p5

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A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodie
vivado [14]

Answer:

The length of the rod for the condition on the question to be met is L  =  1.5077 \ m

Explanation:

The  Diagram for this  question is  gotten from the first uploaded image  

From the question we are told that

          The mass of the rod is M  =  3.41 \ kg

           The mass of each small bodies is  m =  0.249 \ kg

           The moment of inertia of the three-body system with respect to the described axis is   I  =  0.929 \ kg \cdot  m^2

             The length of the rod is  L  

     Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

        I =  I_r + 2 I_m

Where  I_r is the moment of inertia of the rod about the describe axis which is mathematically represented as  

        I_r  =  \frac{ML^2 }{12}

And   I_m the  moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented  as

           I_m  =   m * [\frac{L} {2} ]^2 =  m*  \frac{L^2}{4}

Thus  2 *  I_m  =  2 *  m  \frac{L^2}{4}  = m  *  \frac{L^2}{2}

Hence

       I  =  M  *   \frac{L^2}{12}  +  m  * \frac{L^2}{2}

=>   I  =    [\frac{M}{12}  + \frac{m}{2}] L^2

substituting vales  we have  

        0.929   =    [\frac{3.41}{12}  + \frac{0.249}{2}] L^2

       L  =  \sqrt{\frac{0.929}{0.40867} }

      L  =  1.5077 \ m

     

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