In the periodic table on the inside front cover of this book (or one you find on the internet), for each element there is given the "atomic number," the number of protons or electrons in an atom, and the "atomic mass," which is essentially the number of nucleons, protons plus neutrons, in the nucleus, averaged over the various isotopes of the element, which differ in the number of neutrons. Make a graph of the number of neutrons vs. the number of protons in the elements. You needn't graph every element, just enough to see the trend. What do you observe about the data? (This reflects the need for more neutrons in proton-rich nuclei in order to prevent the electric repulsion of the protons of each other from destroying the nucleus.)

Short Answer

Expert verified

It is evident from the graph that number of neutrons increase as the number of protons increase.

Step by step solution

01

Concept of atomic mass

With each element, the atomic number rises. The mass of an atom is equal to the sum of its protons and neutrons. As the atomic number grows, the atomic mass increases as well.However, the number of neutrons does not always rise in lockstep with the increase in the number of protons.

02

Drawing the required Graph

Draw a graph between number of neutrons and number of protons.

03

Observation from the graph

From the graph, it is clearly visible that as the number of protons increases, number of neutrons also increase along with them. It happens in order to increase the nuclear force so that it can overcome the repulsive electrostatic field caused by protons.

Thus, graph has been drawn and observations have been noted from it.

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

The first stage of giant Saturn V rocket reached a speed of 2300 m/s at 170 s after lift-off.

(a) What was the average acceleration in m/s/s?

(b) The acceleration of a falling object if air resistance is negligible is 9.8 m/s/s, called “one g”. What was the average acceleration in g’s?

Question: Which of the following statements about the velocity and momentum of an object are correct? (1) The momentum of an object is always in the same direction as its velocity (2) The momentum of an object can be either in the same direction as its velocity or in the opposite direction. (3) The momentum of an object is perpendicular to its velocity. (4) The direction of an object's momentum is not related to the direction of its velocity. (5) The direction of an object's momentum is tangent to its path.

In Figurethree vectors are represented by arrows in the xyplane. Each square in the grid represents one meter. For each vector, write out the components of the vector, and calculate the magnitude of the vector.

Figure 1.56

At 6 safter 3:00, a butterfly is observed leaving a flower whose location is<6,-3,10>mrelative to an origin on top of a nearby tree. The butterfly flies until 10 safter 3:00, when it alights on a different flower whose location is<6.8,-4.2,11.2>m relative to the same origin. What was the location of the butterfly at a time 8.5 safter 3:00? What assumption did you have to make in calculating this location?

Figure 1.55 shows several arrows representing vectors in the xy plane. (a) Which vectors have magnitudes equal to the magnitude of a? (b) Which vectors are equal to a?

See all solutions

Recommended explanations on Physics 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