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The image shows wave function curve and potential barrier quantum tunnel region. When the wave function curve passes through potential barrier it decreases exponentially.
The wave function representing a quantum mechanical particle must vary smoothly, going from within the nucleus (to the left of the barrier) to outside the nucleus (to the right of the barrier). Inside the barrier, the wave function does not abruptly become zero; rather, it decreases exponentially. Outside the barrier, the wave function is small but finite, and there it smoothly becomes sinusoidal. Owing to the fact that there is a small probability of finding the particle outside the barrier, the particle can tunnel through the barrier.

Good ideas explain more than one thing. In addition to qualitatively explaining how the four nucleons in an α size 12{α} {} particle can get out of the nucleus, the detailed theory also explains quantitatively the half-life of various nuclei that undergo α size 12{α} {} decay. This description is what Gamow and others devised, and it works for α size 12{α} {} decay half-lives that vary by 17 orders of magnitude. Experiments have shown that the more energetic the α size 12{α} {} decay of a particular nuclide is, the shorter is its half-life. Tunneling explains this in the following manner: For the decay to be more energetic, the nucleons must have more energy in the nucleus and should be able to ascend a little closer to the rim. The barrier is therefore not as thick for more energetic decay, and the exponential decrease of the wave function inside the barrier is not as great. Thus the probability of finding the particle outside the barrier is greater, and the half-life is shorter.

Tunneling as an effect also occurs in quantum mechanical systems other than nuclei. Electrons trapped in solids can tunnel from one object to another if the barrier between the objects is thin enough. The process is the same in principle as described for α size 12{α} {} decay. It is far more likely for a thin barrier than a thick one. Scanning tunneling electron microscopes function on this principle. The current of electrons that travels between a probe and a sample tunnels through a barrier and is very sensitive to its thickness, allowing detection of individual atoms as shown in [link] .

Figure a shows a wavy surface with a taper cylindrical probe traveling horizontally toward right. Wherever there is a crest the barrier between probe and surface is thick and the barrier is thin when there is a hill on the surface. Figure b shows a scanning tunneling electron microscope image of an insect.
(a) A scanning tunneling electron microscope can detect extremely small variations in dimensions, such as individual atoms. Electrons tunnel quantum mechanically between the probe and the sample. The probability of tunneling is extremely sensitive to barrier thickness, so that the electron current is a sensitive indicator of surface features. (b) Head and mouthparts of Coleoptera Chrysomelidea as seen through an electron microscope (credit: Louisa Howard, Dartmouth College)

Phet explorations: quantum tunneling and wave packets

Watch quantum "particles" tunnel through barriers. Explore the properties of the wave functions that describe these particles.

Quantum Tunneling and Wave Packets

Section summary

  • Tunneling is a quantum mechanical process of potential energy barrier penetration. The concept was first applied to explain α size 12{α} {} decay, but tunneling is found to occur in other quantum mechanical systems.

Conceptual questions

A physics student caught breaking conservation laws is imprisoned. She leans against the cell wall hoping to tunnel out quantum mechanically. Explain why her chances are negligible. (This is so in any classical situation.)

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When a nucleus α decays, does the α particle move continuously from inside the nucleus to outside? That is, does it travel each point along an imaginary line from inside to out? Explain.

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Problems-exercises

Derive an approximate relationship between the energy of α decay and half-life using the following data. It may be useful to graph the log of t 1/2 against E α to find some straight-line relationship.

Energy and half-life for α size 12{α} {} Decay
Nuclide E α (MeV) t 1/2
216 Ra 9.5 0.18 μs
194 Po 7.0 0.7 s
240 Cm 6.4 27 d
226 Ra 4.91 1600 y
232 Th 4.1 1.4 × 10 10 y
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Integrated Concepts

A 2.00-T magnetic field is applied perpendicular to the path of charged particles in a bubble chamber. What is the radius of curvature of the path of a 10 MeV proton in this field? Neglect any slowing along its path.

22.8 cm

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(a) Write the decay equation for the α decay of 235 U . (b) What energy is released in this decay? The mass of the daughter nuclide is 231.036298 u. (c) Assuming the residual nucleus is formed in its ground state, how much energy goes to the α particle?

(a) 92 235 U 143 90 231 Th 141 + 2 4 He 2

(b) 4.679 MeV

(c) 4.599 MeV

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Unreasonable Results

The relatively scarce naturally occurring calcium isotope 48 Ca has a half-life of about 2 × 10 16 y . (a) A small sample of this isotope is labeled as having an activity of 1.0 Ci. What is the mass of the 48 Ca in the sample? (b) What is unreasonable about this result? (c) What assumption is responsible?

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Unreasonable Results

A physicist scatters γ rays from a substance and sees evidence of a nucleus 7.5 × 10 –13 m in radius. (a) Find the atomic mass of such a nucleus. (b) What is unreasonable about this result? (c) What is unreasonable about the assumption?

a) 2.4 × 10 8 u

(b) The greatest known atomic masses are about 260. This result found in (a) is extremely large.

(c) The assumed radius is much too large to be reasonable.

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Unreasonable Results

A frazzled theoretical physicist reckons that all conservation laws are obeyed in the decay of a proton into a neutron, positron, and neutrino (as in β + decay of a nucleus) and sends a paper to a journal to announce the reaction as a possible end of the universe due to the spontaneous decay of protons. (a) What energy is released in this decay? (b) What is unreasonable about this result? (c) What assumption is responsible?

(a) –1.805 MeV

(b) Negative energy implies energy input is necessary and the reaction cannot be spontaneous.

(c) Although all conversation laws are obeyed, energy must be supplied, so the assumption of spontaneous decay is incorrect.

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Construct Your Own Problem

Consider the decay of radioactive substances in the Earth’s interior. The energy emitted is converted to thermal energy that reaches the earth’s surface and is radiated away into cold dark space. Construct a problem in which you estimate the activity in a cubic meter of earth rock? And then calculate the power generated. Calculate how much power must cross each square meter of the Earth’s surface if the power is dissipated at the same rate as it is generated. Among the things to consider are the activity per cubic meter, the energy per decay, and the size of the Earth.

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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