How much energy is needed to assemble the Shanghai Tower? Book pg 284 - 285

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Review

• Mass defect is the difference a nucleus and the sum of the mass of its constituent nucleons.

• The mass of a nucleus is always less than the sum of the mass of its constituent nucleons

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Nuclear BE…

• BE is equal to the mass defect

• …is the energy that must be supplied to break a nucleus into separate non interacting nucleons

• BE is a measure of the stability of a nucleus

• Since work must be done to separate nucleons from the nucleus, they must be in a higher energy state

• Nucleons in the nucleus have less energy, hence less mass than free nucleons

• Rest mass of the nucleus is less than sum of rest mass of all nucleons it contains

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Example • Calculate the BE of an uranium – 238 nucleus

𝑈238 = 238.05082u

𝑚𝑒= 0.00055u 𝑚𝑝= 1 00728u 𝑚𝑛= 1 00867u

• #p’s: 92

• #e’s: 92

• #n’s: 238-92=146

• Mass of nucleons: 92 x 1.00728 + 92 x 0.00055 + 146 x 1.00867 = 239.98618u

• ∆𝑚 = 239.98618 − 238.05082 = 1.93536𝑢

• The BE equivalent to the mass defect is BE = 1.93536 x 931.5MeV = 1802.78784 ~ 1803MeV

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Example • Assemble a Helium nucleus from its individual

nucleons and find its BE

• 𝑚𝑒= 0.00055u 𝑚𝑝= 1 00728u = 938.2𝑀𝑒𝑉𝑐−2 𝑚𝑛= 1 00867u = 939.6𝑀𝑒𝑉𝑐−2

𝐻𝑒4 = 4.00260u

• BE of 𝐻𝑒 = 4.00260 × 931.5𝑀𝑒𝑉𝑐−2 = 3728.421924

𝑀𝑒𝑉𝑐−2 ~ 3728.4𝑀𝑒𝑉𝑐−2

• 2 𝑚𝑝 + 2 𝑚𝑛 𝐻𝑒4 + ∆𝑚

• ∆𝑚 = 3755.6 – 3728.4 = 27.2 𝑀𝑒𝑉𝑐−2

• BE = 27.2 MeV

• This means when a helium atom is assembled from nucleons, 28MeV of energy is released

• This means also, 28MeV of energy is required to separate the nucleus into its individual nucleons

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Definition

• Energy required to separate nucleus

• Energy released to assemble nucleus

BE

Remember love L: released - assemble

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PE considerations

• PE nucleus < PE of separate nucleons

• The greater the energy required to separate a nucleus into its nucleons, the greater the difference between the PE of the nucleus and its individual nucleons

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BE per nucleon • To compare stability of different nuclei it is more helpful to use

BE per nucleon than total BE

• BE per nucleon = 𝐵𝐸(𝐸)

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑛𝑢𝑐𝑙𝑒𝑜𝑛𝑠 (𝐴)

• BE p.n.= 𝐸

𝐴

• E = binding energy

• A = number of nucleons

• For Uranium -238 BE (E) was 1803MeV

• BE p.n. = 𝐸

𝐴=

1803

238= 7.6𝑀𝑒𝑉 𝑛𝑢𝑐𝑙𝑒𝑜𝑛−1

• A larger value of BE p.n. usually implies a more stable nucleus

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BE and stability

• The BE tells us how difficult it is to remove each nucleon from the nucleus

• The bigger the binding energy per nucleon, the less likely a nucleus will be to want to lose one of its nucleons

• Thus it is more stable, by definition

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Graph of BE p.n. vs nucleon number • Most values for

𝐸

𝐴 lie between 7 and 9 MeV

• Hydrogen -1 1 has zero BE because it has only 1 nucleon

•𝐸

𝐴 increases rapidly and reaches a peak at iron – 56

• Beyond iron – 56 𝐸

𝐴 falls gradually as nuclei gets larger

Most stable nuclei are those with the greatest BE p.n. because more energy required to separate them

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PRACTICE: Which is the most stable element?

SOLUTION:

The bigger the binding energy per nucleon the more stable an element is.

Iron-56 is the most stable element.

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PRACTICE: What is the binding energy of uranium-238?

SOLUTION:

Note that the binding energy per nucleon of 238U is about 7.6 MeV.

But since 238U has 238 nucleons, the binding energy is

(238)(7.6 MeV) = 1800 MeV.

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PRACTICE: What is the mass defect of uranium-238 if assembled from scratch?

SOLUTION:

The mass defect is equal to the binding energy according to E = mc2.

We need only convert 1800 MeV to a mass.

From the Physics data booklet we see that 1.66110-27 kg = 931.5 MeV c

-2 so that

(1800 MeV)(1.66110-27 kg / 931.5 MeV c-2)

= 3.2110-27 kg.

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PRACTICE: 54Fe has a mass of 53.9396 u. A proton (with electron) has a mass of 1.00782 u and a neutron has a mass of 1.00866 u.

(b) Find the binding energy per nucleon of iron-54.

SOLUTION: 54Fe: A = 54, Z = 26, N = 54 – 26 = 28.

The binding energy is BE = 471.5 MeV c-2.

The number of nucleons is A = 54.

Thus the binding energy per nucleon is

𝐸

𝐴=

471.554

= 8.7 MeV nucleon−1.

This is consistent with the graph.

Example

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Nucleon PE

• Nucleons are trapped in the nucleus as in a potential well created by the strong nuclear force

• Like a golf ball in a hole you need energy to lift it out

• If the surface of the ground has zero potential energy, the ball has negative PE in the hole

• Lifting it up decreases its PE from a negative value to zero

• In the hole, the ball is in a bound state and BE = minimum work done to lift the ball out

• PE of each nucleon in the nucleus is negative

• It is equal in magnitude to BE p.n.

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Nucleon PE

E fusion

E fission

• Fission and fusion convert less stable nuclei into more stable nuclei

• Fusion releases more energy p.n., but nuclei involved contain fewer nucleons than those involved in fission

• Fusion releases more energy per kg • Fission releases more energy per atom

Fusion results in ~ 3 – 4 MeV per nucleon, fission releases less than 1MeV per nucleon

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Nuclear fusion and fission • Fusion

• Joining two lighter nuclei to make a heavier nucleus

• This will decrease average nucleon PE and thus increase its BE p.n.

• PE lost is emitted in other forms such as gamma ray photons or KE of particles

• Fission

• Splitting of heavy nucleus to form two smaller nuclei

• This decreases nucleon PE and releases it in other forms

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Example

• Calculate KE in MeV of tritium plus a helium nucleus in the following nuclear reaction:

𝐿𝑖 + 𝑛 → 𝐻 + 𝐻𝑒24

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• Li = 6.015126u

• H = 3.016030u

• He = 4.002604u

• n = 1.008665u

• ∆𝑚 = 6.015126 + 1.008665 − 4.002604 + 3.016030 = 7.023791 − 7.018634 = 0.005157𝑢

• KE = 0.005157 x 931.5MeV = 4.804MeV

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