   Chapter 31, Problem 23CQ

Chapter
Section
Textbook Problem

To obtain the most precise value of BE from the equation BE = [ Z M ( 1 H ) + N m n ] c 2 − m ( A X ) c 2 , we should take into account the binding energy of the electrons in the neutral atoms. Will doing this produce a larger or smaller value for BE? Why is this effect usually negligible?

To determine

That if on considering the binding energy in the neutral atoms, will this lead to produce a larger or smaller value for BE. Also give reason for considering this effect to be negligible.

Explanation

Explanation:

For a system the attraction forces are directly proportional to how tightly the system is bound together and the amount of energy which is required to pull the system apart should be higher for a tightly bound system.

Thus, we can define for a nucleus the energy of binding is the amount of energy needed for completely separate the nucleus into number of protons and neutrons it has.

We can calculate the energy of binding BE of a nucleus from value of its rest mass.

Consider a nuclide XAhas Z protons and N neutrons. So, the mass defect can be calculated as,

Δm=(Zmp+Nmn)mtot

The binding energy can be written as by using Einstein's relationship E=(Δm)c2

BE=(Δm)c2={(Zmp+Nmn)mtot}c2the

Here, all alphabets are in their usual meanings.

If, we add Z electrons to the Z protons then the binding energy of the nuclide XA will be

BE={(Zm( H1 )+Nmn)

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