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 close to 1 (A/Z ≈ 1). This indicates that the number of protons and neutrons in the nucleus is almost equal. As stable isotopes become heavier, the repulsion between protons increases, requiring more neutrons to stabilize the nucleus. Consequently, the A/Z ratio increases (A/Z > 1) for heavier, stable isotopes as the number of neutrons becomes greater than the number of protons.

Step by step solution

01

Introduction to Isotopes

Isotopes are atoms of the same element (having the same number of protons) but having different numbers of neutrons. This results in different mass numbers (sum of protons and neutrons) for the isotopes of the same element. The mass number is represented by A, and the atomic number is represented by Z.
02

Mass to Atomic Number Ratio for Lighter Isotopes

For lighter, stable isotopes (such as Hydrogen, Helium, Lithium, etc.) the ratio between the mass number (A) and the atomic number (Z) is close to 1. This means that for lighter elements, the number of protons and neutrons in the nucleus is almost equal, making them stable. So, for lighter, stable isotopes, the value of A/Z ≈ 1.
03

Change in Mass to Atomic Number Ratio for Heavier Isotopes

As the isotopes become heavier, the repulsion between protons increases, which in turn requires more neutrons to stabilize the nucleus. Hence, the number of neutrons (N) starts becoming greater than the number of protons (Z) as the isotopes get heavier. Consequently, the mass number (A = Z + N) becomes greater than the atomic number (Z), and the A/Z ratio increases, i.e., A/Z > 1 for heavier, stable isotopes. In conclusion, for lighter, stable isotopes, the A/Z ratio is close to 1, whereas this ratio increases as isotopes become heavier due to an increased number of neutrons needed for stability.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When mixtures of gaseous \(\mathrm{H}_{2}\) and gaseous \(\mathrm{Cl}_{2}\) react, a product forms that has the same properties regardless of the relative amounts of \(\mathrm{H}_{2}\) and \(\mathrm{Cl}_{2}\) used. a. How is this result interpreted in terms of the law of definite proportion? b. When a volume of \(\mathrm{H}_{2}\) reacts with an equal volume of \(\mathrm{Cl}_{2}\) at the same temperature and pressure, what volume of product having the formula \(\mathrm{HCl}\) is formed?

When hydrogen is burned in oxygen to form water, the composition of water formed does not depend on the amount of oxygen reacted. Interpret this in terms of the law of definite proportion.

The formulas and common names for several substances are given below. Give the systematic names for these substances. a. sugar of lead \(\mathrm{Pb}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}\) \(\begin{array}{ll}\text { b. blue vitrol } & \mathrm{CuSO}_{4}\end{array}\) c. quicklime \(\mathrm{CaO}\) d. Epsom salts \(\mathrm{MgSO}_{.}\) e. milk of magnesia \(\quad \mathrm{Mg}(\mathrm{OH})_{2}\) f. gypsum \(\mathrm{CaSO}_{4}\) \(\mathrm{g}\). laughing gas \(\mathrm{N}_{2} \mathrm{O}\)

Carbohydrates, a class of compounds containing the elements carbon, hydrogen, and oxygen, were originally thought to contain one water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) for each carbon atom present. The carbohydrate glucose contains six carbon atoms. Write a general formula showing the relative numbers of each type of atom present in glucose.

Dalton assumed that all atoms of the same element were identical in all their properties. Explain why this assumption is not valid.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free