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NeTakaya
3 years ago
5

How does the mass and size of an atom compare to the mass and size of the nucleus

Chemistry
2 answers:
Studentka2010 [4]3 years ago
8 0

Answer:

The nucleus of an atom is about 10-15 m in size; this means it is about 10-5 (or 1/100,000) of the size of the whole atom. ... (10-15 m is typical for the smaller nuclei; larger ones go up to about 10 times that.) Mass. Although it is very small, the nucleus is massive compared to the rest of the atom.

Explanation:

ira [324]3 years ago
6 0

Answer:

Explanation:

The nucleus of an atom is about 10-15 m in size; this means it is about 10-5 (or 1/100,000) of the size of the whole atom. ... (10-15 m is typical for the smaller nuclei; larger ones go up to about 10 times that.) Mass. Although it is very small, the nucleus is massive compared to the rest of the atom.

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An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If t
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Answer:

\boxed{3}

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )

where

R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}

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Calculation:  

\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}

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