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For lighter, stable isotopes, the ratio of the mass number to the atomic number is close to a certain value. What is the value? What happens to the value of the mass number to atomic number ratio as stable isotopes become heavier?

Short Answer

Expert verified
For lighter, stable isotopes, the ratio of the mass number (A) to the atomic number (Z) is approximately 2, meaning \(\frac{A}{Z} \approx 2\). As stable isotopes become heavier, the ratio \(\frac{A}{Z}\) increases due to the need for more neutrons to maintain stability in the nucleus against the stronger electrostatic repulsion caused by a larger number of protons.

Step by step solution

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1. Determining the ratio for lighter, stable isotopes

For lighter, stable isotopes, the ratio of the mass number (A) to the atomic number (Z) is approximately 2. This means that the mass number (number of protons and neutrons) is twice the atomic number (number of protons): \[ \frac{A}{Z} \approx 2 \]
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2. Understanding the trend with heavier isotopes

As stable isotopes become heavier, the ratio \(\frac{A}{Z}\) increases. This is because adding more neutrons is necessary to increase the stability of the nucleus for elements with a higher atomic number (more protons). The larger number of protons leads to a stronger electrostatic repulsion within the nucleus, so adding more neutrons with their strong nuclear force helps offset this repulsion and maintain stability.
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3. Conclusion

The ratio of the mass number to the atomic number for lighter, stable isotopes is approximately 2. As the isotopes become heavier, the value of this ratio increases due to the need for an increased number of neutrons for stability.

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Most popular questions from this chapter

You have two distinct gaseous compounds made from element \(\mathrm{X}\) and element Y. The mass percents are as follows: Compound I: \(30.43 \%\) X, \(69.57 \%\) Y Compound II: \(63.64 \% \mathrm{X}, 36.36 \% \mathrm{Y}\) In their natural standard states, element X and element Y exist as gases. (Monatomic? Diatomic? Triatomic? That is for you to determine.) When you react "gas X" with "gas Y" to make the products, you get the following data (all at the same pressure and temperature): 1 volume "gas \(\mathrm{X}^{\prime \prime}+2\) volumes "gas \(\mathrm{Y}^{\prime \prime} \longrightarrow\) 2 volumes compound I 2 volumes "gas \(\mathrm{X}^{\prime \prime}+1\) volume "gas \(\mathrm{Y}^{\prime \prime} \longrightarrow\) 2 volumes compound II Assume the simplest possible formulas for reactants and products in the chemical equations above. Then, determine the relative atomic masses of element \(X\) and element Y.

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