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DiKsa [7]
3 years ago
12

The discovery of the atomic nucleus, or perhaps better stated the development of theory of the nucleus, is credited to Ernest Ru

therford in 1911. He was a kiwi physicist who worked in the laboratory of JJ. Tomson, who developed the earlier Plum Pudding model of the atom. A few years earlier, in 1904, Japanese physicist Hantaro Nagaoka postulated the saturnian model of the atom where he envisioned a very heavy positively charged center of the atom with electrons orbiting much like the rings of Saturn. What are the similarities and differences between the Nagaoka and Rutherford models of the atom? If he developed his theory about 7 years earlier than Rutherford, why is Nagaoka not credited with the discovery of the nucleus? Hint: what's a nucleus anyways
Chemistry
1 answer:
Greeley [361]3 years ago
3 0

Answer:

Both of the studies said that the mass of the atom is centered in the nucleus, which is positive, and there are electrons (negative particles) orbiting it. So, Rutheford and Nagaoka discovered that the atom can be divisible and it has an empty space.

But, in the model of Nagaoka, the nucleus was huge, and for Rutherford, the nucleus was really small, and the mass was concentrated. By his experiment with the gold sheets, the theory was appropriated. That's why Rutherford is credited with the discovery of the nucleus. Nagaoka was incorrect in his suppositions.

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Explanation:

<em>Ground-state</em> is the configuration of the electrons when every electron of the atom occupies the lowest possible energy level.

The rank of energy of the orbitals may be remembered using Aufbau rules, and it is:

1s² < 2s² < 2p⁶ < 3s² < 3p⁶ < 4s² < 3d¹⁰ < 4p⁶ < 5s² < 4d¹⁰ < 5p⁶ < 6s² < 4f¹⁴ < 5d¹⁰ < 6p⁶ < 7s² < 5f¹⁴ < 6d¹⁰ < 7p⁶

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The orbital 1p does not exist. So this one is discarded.

<u>Option B) 1s² 2s⁴ 2p⁶</u>

The maximum number of electrons in an s orbital is 2, so 2s⁴ is impossible, and this option is discarded.

<u>Option C) 1s² 2s² 2p⁵</u>

This is correct because the orbitals are filled in the correct increasing energy order, without jumping any one.

<u>Option D) 1s² 2s² 2d⁶</u>

Orbital 2d does not exist, so this option is also discarded.

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