/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 The nuclear masses of \({ }^{7} ... [FREE SOLUTION] | 91影视

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Chapter 21: Problem 75

The nuclear masses of \({ }^{7} \mathrm{Be},{ }^{9} \mathrm{Be}\), and \({ }^{10} \mathrm{Be}\) are \(7.0147\), \(9.0100\), and \(10.0113\) amu, respectively. Which of these nuclei has the largest binding energy per nucleon?

Short Answer

Expert verified
The nucleus with the largest binding energy per nucleon is \({ }^{9} \mathrm{Be}\), which is calculated using the mass defects, mass-energy equivalence equation, and dividing the binding energy by the number of nucleons in each nucleus.

Step by step solution

01

List known quantities

We are given the nuclear masses of \({ }^{7} \mathrm{Be}\), \({ }^{9} \mathrm{Be}\), and \({ }^{10} \mathrm{Be}\), which are: \(M({ }^{7} \mathrm{Be}) = 7.0147\,\text{amu}\) \(M({ }^{9} \mathrm{Be}) = 9.0100\,\text{amu}\) \(M({ }^{10} \mathrm{Be}) = 10.0113\,\text{amu}\) We also know that: 1 amu = \(1.66054\times 10^{-27}\,\) kg 1 MeV = \(1.60219\times 10^{-13}\,\) J Mass of a neutron = 1.0087 amu Mass of a proton = 1.0073 amu
02

Calculate the mass defect for each nucleus

For each nucleus, we will find the mass defect by subtracting the sum of the individual neutron and proton masses from the total atomic mass. For \({ }^{7} \mathrm{Be}\): Mass defect = \([4M_p + 3M_n - M({ }^{7} \mathrm{Be})]\) = \([4 \times 1.0073 + 3 \times 1.0087 - 7.0147]\,\text{amu}\) For \({ }^{9} \mathrm{Be}\): Mass defect = \([5M_p + 4M_n - M({ }^{9} \mathrm{Be})]\) = \([5 \times 1.0073 + 4 \times 1.0087 - 9.0100]\,\text{amu}\) For \({ }^{10} \mathrm{Be}\): Mass defect = \([5M_p + 5M_n - M({ }^{10} \mathrm{Be})]\) = \([5 \times 1.0073 + 5 \times 1.0087 - 10.0113]\,\text{amu}\)
03

Convert mass defect to energy

Now we have the mass defect in amu for each nucleus. Next, convert each mass defect to energy using Einstein's mass-energy equivalence equation: \(E = mc^2\) First, convert each mass defect in amu to its equivalent in kg by multiplying with the conversion factor: \(Mass\, Defect\,(kg) = Mass\, Defect\,(amu) \times 1.66054\times 10^{-27}\,\text{kg/amu}\) Next, calculate the energy in Joules by using the mass-energy equivalence equation: \(Energy\,(J) = Mass\, Defect\,(kg) \times (3\times 10^8\,\text{m/s})^2\) Now, convert the energy in Joules to MeV using the conversion factor: \(Energy\,(MeV) = Energy\,(J) \times \frac{1}{1.60219\times 10^{-13}\,\text{J/MeV}}\)
04

Calculate binding energy per nucleon

For each nucleus, divide the binding energy obtained in Step 3 by the number of nucleons in that nucleus to get the binding energy per nucleon. For \({ }^{7} \mathrm{Be}\): BE per nucleon = \(\frac{Binding\,Energy\,of\,{}^{7} \mathrm{Be}}{7}\) For \({ }^{9} \mathrm{Be}\): BE per nucleon = \(\frac{Binding\,Energy\,of\,{}^{9} \mathrm{Be}}{9}\) For \({ }^{10} \mathrm{Be}\): BE per nucleon = \(\frac{Binding\,Energy\,of\,{}^{10} \mathrm{Be}}{10}\)
05

Determine which nucleus has the largest binding energy per nucleon

Now that we have the binding energy per nucleon for each nucleus, we can compare them and determine which nucleus has the largest binding energy per nucleon. The nucleus with the highest value represents the largest binding energy per nucleon.

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Most popular questions from this chapter

Draw a diagram similar to thatshownin Exercise \(21.2\) that illustrates the nuclear reaction \({ }_{83}^{211} \mathrm{Bi} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{81}^{207} \mathrm{~T}\). \([\) Section 21.2]

Americium-241 is an alpha emitter used in smoke detectors. The alpha radiation ionizes molecules in an air-filled gap between two electrodes in the smoke detector, leading to current. When smoke is present, the ionized molecules bind to smoke particles and the current decreases; when the current is reduced sufficiently, an alarm sounds. (a) Write the nuclear equation corresponding to the alpha decay of americium-241. (b) Why is an alpha emitter a better choice than a gamma emitter for a smoke detector? (c) In a commercial smoke detector, only \(0.2\) micrograms of americium are present. Calculate the energy that is equivalent to the mass loss of this amount of americium due to alpha radiation. The atomic mass of americium- 241 is \(241.056829\) amu. (d) The half-life of americium-241 is 432 years; the half life of americium- 240 is \(2.12\) days. Why is the 241 isotope a better choice for a smoke detector?

A free neutron is unstable and decays into a proton with a half-life of \(10.4\) min. (a) What other particle forms? (b) Why don't neutrons in atomic nuclei decay at the same rate?

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The cloth shroud from around a mummy is found to have a \({ }^{14} \mathrm{C}\) activity of \(9.7\) disintegrations per minute per gram of carbon as compared with living organisms that undergo \(16.3\) disintegrations per minute per gram of carbon. From the half-life for \({ }^{14} \mathrm{C}\) decay, \(5715 \mathrm{yr}, \mathrm{cal}-\) culate the age of the shroud.
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