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gladu [14]
3 years ago
6

A neutral atom of which of the four element has the smallest radius

Chemistry
2 answers:
arsen [322]3 years ago
6 0
Answer:
Element 1

Explanation:
The ionization energy is defined as the energy required to remove electrons from the atoms.
We know that the nucleus of the atom attracts the electrons, thus, bound these electrons to the atom.
This means that as the radius decreases, the force of attraction between the nucleus and the electron will increase, therefore, the energy required to remove the electron would increase (and vice-versa).

Based on the above, the atom with the smallest radius would be the atom with the largest first ionization energy.

Hope this help :)
Dennis_Churaev [7]3 years ago
4 0
I would probably say that element 1 has the lowest atomic radius since it has the highest 1st ionization energy.
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3 years ago
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Answer:

P = 1/8

Explanation:

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P = \int_{x_{1}}^{x_{2}} \psi^{2} dx

P = \int_{x_{1}}^{x_{2}} (\sqrt \frac{2}{L} sin(\frac{n \pi x}{L}))^{2} dx

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Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:

P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]

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Solving for n=4:

P = \frac{1}{8} (1 - 4\frac{sin(\frac{4 \pi}{4})}{4 \pi})    

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I hope it helps you!

7 0
3 years ago
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