Nuclear binding energy and binding energy per nucleon for the isotope of uranium has to be calculated for U 238 which has an isotopic mass of 238 .0508 g/mol . Concept Introduction: Binding energy is a short strong force that is present in the nucleus which holds the protons together by overcoming the electrostatic repulsive forces between them. Whenever there is a change in energy, a corresponding change in mass is also observed and this can be given by the equation shown below, Δ E = ( Δ m ) c 2 When more particles combine to form nuclear there is a great change in mass and energy. The nuclear stabilities can be compared more appropriately by dividing the binding energy of nucleus with the number of nucleons. The result obtained is the binding energy per nucleon. Protons and neutrons are known as nucleons. Binding energy is represented as E b .

Question

Want to see the full answer?

Answer 1

Question

Chapter 18, Problem 25QRT

Interpretation Introduction

Interpretation:

Nuclear binding energy and binding energy per nucleon for the isotope of uranium has to be calculated for U238 which has an isotopic mass of 238.0508 g/mol.

Concept Introduction:

Binding energy is a short strong force that is present in the nucleus which holds the protons together by overcoming the electrostatic repulsive forces between them. Whenever there is a change in energy, a corresponding change in mass is also observed and this can be given by the equation shown below,

ΔE = (Δm)c2

When more particles combine to form nuclear there is a great change in mass and energy. The nuclear stabilities can be compared more appropriately by dividing the binding energy of nucleus with the number of nucleons. The result obtained is the binding energy per nucleon. Protons and neutrons are known as nucleons. Binding energy is represented as Eb.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

A J то گای ه +0 Also calculate the amount of starting materials chlorobenzaldehyde and p-chloroacetophenone required to prepare 400 mg of the given chalcone product 1, 3-bis(4-chlorophenyl)prop-2-en-1-one molar mass ok 1,3-bis(4-Chlorophenyl) prop-2-en-1-one = 277.1591m01 number of moles= 0.400/277.15 = 0.00144 moles 2 x 0.00 144=0.00288 moves arams of acetophenone = 0.00144 X 120.16 = 0.1739 0.1739x2=0.3469 grams of benzaldehyde = 0.00144X106.12=0.1539 0.1539x2 = 0.3069 Starting materials: 0.3469 Ox acetophenone, 0.3069 of benzaldehyde 3

1. Answer the questions about the following reaction: (a) Draw in the arrows that can be used make this reaction occur and draw in the product of substitution in this reaction. Be sure to include any relevant stereochemistry in the product structure. + SK F Br + (b) In which solvent would this reaction proceed the fastest (Circle one) Methanol Acetone (c) Imagine that you are working for a chemical company and it was your job to perform a similar reaction to the one above, with the exception of the S atom in this reaction being replaced by an O atom. During the reaction, you observe the formation of three separate molecules instead of the single molecule obtained above. What is the likeliest other products that are formed? Draw them in the box provided.

3. For the reactions below, draw the arrows corresponding to the transformations and draw in the boxes the reactants or products as indicated. Note: Part A should have arrows drawn going from the reactants to the middle structure and the arrows on the middle structure that would yield the final structure. For part B, you will need to draw in the reactant before being able to draw the arrows corresponding to product formation. A. B. Rearrangement ΘΗ