Radioactivity and energy levels - WordPress.com

Radioactivity and energy levels

Book page 497 - 503 ©cgrahamphysics.com 2016

Review of radioactivity

𝜷- decay is continuous • 𝛽;

• Free neutron proton

• antineutrino

• 𝛽:

• Proton in nucleus neutron

• neutrino

©cgrahamphysics.com 2016

Summary of useful equations

• Rate of decay: ⅆ𝑁

ⅆ𝑡= −λ𝑁

• Activity: A = −ⅆ𝑁

ⅆ𝑡= λ𝑁 or just A = λ𝑁

𝐴 =𝐴0

2𝑛

• Decay equation: 𝑁 = 𝑁0𝑒;λ𝑡

𝑁 = 𝑁0

1

2

𝑛 𝑤𝑕𝑒𝑟𝑒 𝑛 =

∆𝑡

𝑇12

• Half life: 𝑇1

2

=ln 2

λ

• Fraction remaining after time t: 𝑒;λ𝑡

This means 𝑒;λ𝑇1

2 =1

2


Definitions Half life

• The time it takes the activity of a sample of the element to half in value

• Or

• The time it takes for half of the atoms in the sample of the element to decay

• Do NOT use mass

Decay constant

• The probability of decay of a nucleus per unit time

• Random and unpredictable

• The more radioactive nuclei are present the greater the probability of some decaying

Radioactive decay


Example • Radium – 226 emits alpha particles. The decay

constant is 1.35 × 10;11. What mass of radium 226 is needed to give an activity of 2200Bq?

Solution

• A = λ𝑁

• 𝑁 =𝐴

λ=

2200

1.35×10−11= 1.63 × 1014

• 𝑛 =𝑁

𝑁𝐴=

1.63×1014

6.02×1023= 2.7 × 10;10𝑚𝑜𝑙

• 𝑚 = 𝑛𝑀 = 226 × 10;3 × 2.7 × 10;10 = 6.1 × 10;11𝑘𝑔


Example • A laboratory prepares a 10𝜇g sample of caesium -134. The half life of

caesium – 134 is about 2.1 years. a) determine, in seconds, the decay constant for this isotope b) calculate the initial activity of the sample c) calculate the activity of the sample after 10.0 years

Solution

• A) 𝑇1

2

=ln 2

λ λ =

ln 2

𝑇12

=0.693

2.1×365×24×3600= 1.05 × 10;8𝑠;1

• B) 𝑛 =𝑚

𝑀=

10×10−6

134= 7.5 × 10;8𝑚𝑜𝑙

𝑁 = 𝑛𝑁𝐴 = 7.5 × 10;8 ×6.02 × 1023 = 4.49 × 1016 𝐴 = λ𝑁 = 1.05 × 10;8 × 4.49 × 1016 = 4.7 × 108𝐵𝑞

• C) 𝐴 = 𝐴0𝑒;λ𝑡 = 4.7 × 108𝑒;0.33×10 = 1.7 × 107𝐵𝑞

or 0.33 years


Measuring radioactive half life How to measure depends on the length of the half life

Long half life

• Known mass of sample

• Measure activity using Geiger counter

• Use decay equation

• Find decay constant

Example

• A sample of isotope of uranium -234 has a mass of 2.0 𝜇g. Its activity is measured 3.0 × 103𝐵𝑞. What is its half life?

Solution

• Given: A = ⅆ𝑁

ⅆ𝑡= 3.0 × 103Bq

• 𝑁 =𝑚

𝑁𝐴=

2.0×10−6

6.02×1023= 3.3 × 1016 atoms

•ⅆ𝑁

ⅆ𝑡= −λ𝑁

λ=

ⅆ𝑁

ⅆ𝑡

𝑁=

3.0×103

3.3×1016= 9.0 × 10;14𝑠;1

𝑇12=ln 2

λ=

ln 2

9.0 × 10;14𝑠;1= 7.6 × 1012𝑠 = 2.4 × 105 𝑦𝑒𝑎𝑟𝑠


Very long half life • Geiger counter cannot be used for very long 𝑇1

2

such as uranium –

238, which has a half life of just under 4.5 billion years • The rate of decay cannot be measured • A pure sample of nuclide in a known chemical form needs to be

separated, its mass measured and then a count rate taken • From the reading the activity can be calculated by multiplying the

count rate by the ratio of 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑝𝑕𝑒𝑟𝑒 𝑜𝑓 𝑟𝑎𝑑𝑖𝑢𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐺 − 𝑀 𝑡𝑢𝑏𝑒 𝑤𝑖𝑛𝑑𝑜𝑤

𝑎𝑟𝑒𝑎 𝑜𝑓 𝐺 − 𝑀 𝑡𝑢𝑏𝑒 𝑤𝑖𝑛𝑑𝑜𝑤

• Determine decay constant from mass of specimen using the same method as for long half life


Short half life – order of hours

• Measure number of decays in short period (minutes) at different time intervals

• Plot A versa time and find 𝑇1

2

from the graph

• Plot log A versa time

• Yield a straight line

• Gradient = −λ Gradient = - λ

Log A

time


Very short half life – order of seconds • Use ionization properties

• Place sample in tube

• Apply electric field across tube

• Radiation from source will ionize air

• Ionization current is set up

• Decay of ionization current can be displayed on an oscilloscope


Nuclear energy levels • Radioactive decay provides evidence for nucleus having

energy levels

• Daughter nucleus emits one or more 𝛾 ray photons as it reaches ground state

• Emission of 𝛼 𝑜𝑟 𝛽 particles by radioactive parent nuclei often leaves daughter nucleus in excited state


𝛼- decay • Decay route of Americum – 241 to Neptunium - 237

• Each nucleus emits an alpha particle having one of three possible energies (there are more, but for this example we have 3)

• Depending on emitted 𝛼 particle Neptunium can be in ground state or in an excited state

• It will decay into ground state by emitting a single photon or, in two steps, two photons

• 𝛼 particles have quantized energies

• 𝛾 particles have also quantized energies

• Hence nucleus must have energy levels ©cgrahamphysics.com 2016

Mechanism of 𝛼 decay leaving nucleus • 𝛼 particles form as clusters 2p + 2n inside the nucleus

• Nucleons are in random motion but energy less than energy needed to escape the nucleus

• Strong nuclear force provides potential energy barrier

• 𝛼 particles need to overcome this (electrostatic force will then accelerate it away from the nucleus)

• According to classical mechanics the alpha particles do not have enough energy to leave


𝛼 decay and quantum mechanics

• Ψ of 𝛼 particles not localized to nucleus

• Allows overlap with potential barrier provided by strong nuclear force

• There is a finite but very small probability of observing the alpha particle outside the nucleus

• Some particles will tunnel out of the nucleus


Experiments show

• With a higher potential barrier and a greater thickness to cross, a nucleus will have a larger life time

• This explains very long half lives of uranium and polonium

• When the wave function is at its maximum the probability of tunneling is greatest

• 𝛼 particles with specific energies are most likely to be emitted


𝛽 decay is continuous

• It is continuous because neutrino accounts for any energy difference between maxima 𝛽 decay and the sum of 𝛾 plus intermediate 𝛽 particle energies

• Pauli suggested that a third particle was emitted in the decay to conserve spin and angular momentum as well as mass – energy conservation and momentum


Fictinous example of 𝛽: decay

• 𝛽; decay neutron proton + 𝑣

• 𝛽: decay proton neutron + 𝑣

• The neutrino accounts for the continuous 𝛽 spectrum


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