# Quantum Tunneling Probability

> A quantum particle can sometimes pass through a barrier that is too high for it to climb over, and this formula gives a rough estimate of how often.

Page: https://formulabro.yunusemrevurgun.com/quantum-mechanics/quantum-tunneling-probability/  
Category: Quantum Mechanics  
Published: 2026-10-22  
Updated: 2026-10-22

## Formula

$$
T \approx e^{-2\kappa L}
$$

Read it out loud: "T is approximately equal to e to the power of minus 2 kappa L. T is the chance that the particle gets through the barrier, a number from 0 to 1. e is Euler's number, about 2.718. Kappa (KAP-uh) is how fast the wave fades inside the barrier. L is the width of the barrier. Minus 2 kappa L is the exponent: two times kappa times L, with a minus sign in front."

## Before you start

- **Powers of ten, like $3.0 \times 10^{-10}$ m**: The example uses numbers such as a barrier width of $3.0 \times 10^{-10}$ m and an electron mass of $9.1093837 \times 10^{-31}$ kg. Each power of ten says how far to move the decimal point. The exponent $-10$ moves it ten places to the left, so $3.0 \times 10^{-10}$ m is 0.0000000003 m, which is 0.3 nanometers. To multiply powers of ten, add the exponents: $10^{-10} \times 10^{10} = 10^{0} = 1$.
- **Euler's number $e$ and the exponential $e^{x}$**: The letter $e$ here stands for **Euler's number**, about 2.71828, a fixed number from mathematics. It appears whenever something shrinks or grows in proportion to its own size. A negative power such as $e^{-5}$ means one divided by $e^{5}$, which is a very small number, about 0.0067. The more negative the exponent, the smaller the result. This letter is not the electron's charge.
- **Probability as a chance from 0 to 1**: A **probability** is a chance written as a number from 0 to 1. Zero means the event never happens, and 1 means it always happens. A **percent** is a part out of one hundred, so a probability of 0.0058 is 0.58 percent. That means about 58 times in 10,000 tries. The probabilities of all possible outcomes add up to 1.
- **The Greek letter $\kappa$ (kappa)**: The Greek letter $\kappa$ is read KAP-uh. In this formula it is the **decay constant**, which says how fast the wave dies away as it enters the barrier. Its unit is 1/m, one per meter. So $\kappa L$ has no unit, because one per meter times meters cancels. A large $\kappa$ means the wave fades quickly, so the particle has a small chance of getting through.
- **Potential energy and the barrier height $V_0$**: **Potential energy** is energy stored because of position, like a ball resting at the top of a hill. A **barrier** is a region where the potential energy is high. Its height is written $V_0$, the potential energy at the top of the barrier, measured in joules or electron volts. A particle with energy $E$ below $V_0$ does not have enough energy to climb over the top in ordinary physics.
- **Electron volts and the joule**: An **electron volt** (eV) is a small unit of energy, the energy an electron gains when pushed through one volt. It equals $1.602176634 \times 10^{-19}$ J. The **joule** (J) is much larger: it is roughly the energy needed to lift a 100 gram apple about one meter. Atomic energies are written in electron volts, because joules would give very long decimal numbers.

More notation help: https://formulabro.yunusemrevurgun.com/math-basics/

## What each symbol means

- $T$: The **transmission probability**, the chance that a particle gets through the barrier. It is a number from 0 to 1 and has no unit. A value of 0.0058 means about 0.58 percent of the particles get through, on average.
- $e$: **Euler's number**, about 2.71828, a fixed number from mathematics and the base of natural logarithms. It has no unit. This letter is not the elementary charge of the electron, although that charge is also written $e$ in other places.
- $\kappa$: The **decay constant** inside the barrier, in 1/m (per meter). It equals $\kappa = \frac{\sqrt{2m(V_0-E)}}{\hbar}$. A large $\kappa$ means the wave dies out quickly as it enters the barrier, so the chance of getting through is small.
- $L$: The **width of the barrier**, the distance the particle would have to cross, in meters (m). In the example, $L = 0.30$ nm $= 3.0 \times 10^{-10}$ m. Doubling the width squares the chance, which makes it much smaller.
- $m$: The **mass** of the particle, in kilograms (kg). For an electron, $m = 9.1093837 \times 10^{-31}$ kg. A heavier particle has a larger $\kappa$, so it tunnels less often.
- $E$: The **energy** of the particle, in joules (J), or in electron volts when that is more convenient. The formula assumes $E < V_0$, so the particle does not have enough energy to climb over the barrier. In the example, $E = 0.20$ eV.
- $V_0$: The **height** of the barrier, the potential energy at its top, in joules (J). In the example, $V_0 = 3.0$ eV, which is $4.8 \times 10^{-19}$ J, because one electron volt is $1.602 \times 10^{-19}$ J.
- $\hbar$: The **reduced Planck constant**, $1.054571817 \times 10^{-34}$ J s, which is Planck's constant divided by $2\pi$. It sets how strongly quantum effects show up. A smaller $\hbar$ would make $\kappa$ larger.

## The formula piece by piece

- $V_0 - E$: Start with the missing energy. $V_0 - E$ is the barrier height minus the particle's energy, which is how much energy the particle lacks to climb over the top. In the example, $3.0$ eV $- 0.20$ eV $= 2.8$ eV. A bigger gap makes the wave fade faster.
- $\kappa = \frac{\sqrt{2m(V_0-E)}}{\hbar}$: Next, $\kappa$ is found from the missing energy, the mass and $\hbar$. The square root of $2m(V_0 - E)$ grows with both the mass and the missing energy, and dividing by $\hbar$ turns it into a fade rate in 1/m. Heavy particles and tall barriers therefore tunnel less.
- $2\kappa L$: Multiply the fade rate by the width and by 2. The product $\kappa L$ counts how many fade lengths fit across the barrier. The factor 2 comes from the probability, which is the wave squared. In the example, $2\kappa L \approx 5.14$.
- $e^{-2\kappa L}$: Use $e$ raised to the power of minus $2\kappa L$. The minus sign means the chance shrinks exponentially as the barrier gets wider or the fade rate gets bigger. In the example this gives a number of about 0.0058.
- $T \approx e^{-2\kappa L}$: The whole formula in one sentence: **the chance of getting through the barrier is about $e$ raised to the power of minus two times the fade rate times the width.**

## What the formula is doing, and why

In everyday life, a ball that lacks the energy to roll over a hill stays on its side. In quantum mechanics, the wave of a particle does not stop sharply at the edge of a barrier. It fades inside the barrier, and a small part of it can reach the far side. What goes in is the energy, the barrier height and width, and the mass. What comes out is the chance $T$ that the particle gets through.

**Why the missing energy and the mass?** The energy $E$ is the particle's energy and $V_0$ is the barrier height, so $V_0 - E$ is the energy the particle is missing. The fade rate $\kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar}$ grows with the square root of the mass and with the missing energy. Heavy particles and tall barriers make the wave fade faster, so they tunnel less.

**Why the exponential?** The chance falls exponentially with the width $L$. Each extra bit of width multiplies $T$ by the same fraction, rather than subtracting the same amount. In this estimate, doubling the width squares $T$, which makes the chance drop very fast. The shape comes from the wave fading as $e^{-\kappa x}$ inside the barrier, and the probability is the wave squared, which doubles the exponent.

**Does the particle climb over?** No. Tunneling is not a way of climbing over the barrier. The particle keeps its energy $E$ and does not borrow energy. The result is a chance, and most attempts fail, so the particle is usually reflected. Quantum mechanics says the particle has a small chance of appearing on the far side, and no sudden jump of energy is needed.

**How good is the estimate?** This is a thick-barrier estimate. The fuller thick-barrier result carries an extra prefactor, $\frac{16E}{V_0}\left(1 - \frac{E}{V_0}\right)$, which the simple formula drops. In the example below this prefactor is about 0.996, so it hardly changes the answer. It is not always close to 1. It equals 4 when $E$ is half of $V_0$, and it is much less than 1 when $E$ is far below $V_0$.

## An everyday picture

Think of the sound of a neighbor's music through a wall. A thin wall lets through plenty of bass, while a thick wall removes much more. Each extra bit of thickness takes away the same fraction of what is left, though a little always gets through. Tunneling works the same way: the wave fades exponentially inside the barrier, so a thicker barrier lets far less through, but never exactly nothing.

## Worked example: An electron meets a 3.0 eV barrier

An electron with energy $E = 0.20$ eV meets a barrier of height $V_0 = 3.0$ eV and width $L = 0.30$ nm $= 3.0 \times 10^{-10}$ m. The electron mass is $m = 9.1093837 \times 10^{-31}$ kg. We estimate the chance $T$ that the electron tunnels through, and then how the chance changes if the barrier is twice as wide.

1. **Find the missing energy.** The barrier is 3.0 eV and the electron has 0.20 eV, so the missing energy is $V_0 - E = 3.0 - 0.20 = 2.8$ eV. The formula needs joules, so convert with $1\ \text{eV} = 1.602 \times 10^{-19}$ J: $2.8 \times 1.602 \times 10^{-19} = 4.486 \times 10^{-19}$ J.
2. **Compute the fade rate $\kappa$.** Multiply $2m(V_0 - E)$: $2 \times 9.1093837 \times 10^{-31} \times 4.486 \times 10^{-19} \approx 8.173 \times 10^{-49}$ kg J. For the square root, write this as $81.73 \times 10^{-50}$ so the exponent is even, then $\sqrt{81.73 \times 10^{-50}} \approx 9.041 \times 10^{-25}$. Divide by $\hbar$: $\kappa \approx \frac{9.041 \times 10^{-25}}{1.0546 \times 10^{-34}} \approx 8.573 \times 10^{9}$ m$^{-1}$.
3. **Multiply the fade rate by the width.** $\kappa L = (8.573 \times 10^{9}\ \text{m}^{-1})(3.0 \times 10^{-10}\ \text{m})$. The front numbers give $8.573 \times 3.0 = 25.72$, and the powers of ten give $10^{9} \times 10^{-10} = 10^{-1}$. So $\kappa L \approx 2.572$, with the meters canceling.
4. **Double it to get the exponent.** The exponent is $2\kappa L = 2 \times 2.572 = 5.144$. The factor 2 is there because the chance is the wave squared, so the fade is counted twice.
5. **Estimate the chance.** $T \approx e^{-5.144}$. A calculator gives about 0.0058, which is 0.58 percent. In words, roughly 1 particle in 170 gets through this barrier, which is small but not zero.
6. **Check the dropped prefactor.** The fuller result multiplies by $\frac{16E}{V_0}\left(1 - \frac{E}{V_0}\right) = 16 \times \frac{0.20}{3.0} \times \frac{2.8}{3.0} \approx 0.996$. That changes the answer very little. A complete calculation for this rectangular barrier gives 0.584 percent, which agrees with the estimate to less than 0.2 percent.
7. **Double the width.** Doubling $L$ doubles the exponent, so the chance becomes $T$ squared. In this estimate the chance is about 171 times smaller: $e^{2\kappa L} = e^{5.144} \approx 171$. Going from 0.30 nm to 0.60 nm makes tunneling far less likely.

**Answer:** The chance that the electron tunnels is about **0.58 percent**, or roughly **1 in 170**. That is small but not zero. Doubling the barrier width to 0.60 nm would make the chance about **171 times smaller**.

## Common mistakes

- Thinking a particle with less energy than the barrier can never get through. Quantum particles have a small chance of tunneling, and the formula estimates that chance.
- Using the barrier height $V_0$ where the formula needs the missing energy $V_0 - E$. Subtract first, then take the square root.
- Leaving the energies in electron volts. The formula needs joules, so convert $V_0 - E$ with $1\ \text{eV} = 1.602 \times 10^{-19}$ J. Skipping this makes $\kappa$ about $2.5 \times 10^{9}$ times too large.
- Treating the answer as exact. The formula drops a prefactor, so it estimates the exponential part only. For some energies it can be off by a factor of a few.

## Where you will meet it

- Scanning tunneling microscopes, which pass a tiny current across a gap of less than a nanometer. The current depends steeply on the gap, which lets the microscope image single atoms.
- Alpha decay, where an alpha particle escapes a heavy nucleus by tunneling through the energy barrier that holds it inside.
- Nuclear fusion in the Sun, where protons must tunnel through their electrical repulsion to get close enough to fuse.
- Flash memory, where electrons tunnel through a thin insulating oxide layer to store or erase data.

## Questions people ask

### Does tunneling mean the particle borrows energy?

No. In this picture the particle keeps its energy $E$. Tunneling is not a way of climbing over the barrier. The wave has a small chance of reaching the far side.

### Why does a thicker barrier block so much more?

The chance depends on $e^{-2\kappa L}$, so every extra bit of width multiplies the chance by the same factor. In the example, $2\kappa L$ grows by about 1.7 for each extra 0.1 nm, which cuts the chance to about 0.18 of its value.

### Is the formula exact?

No. It is a thick-barrier estimate, and it drops a prefactor that a fuller calculation contains. For the example here, the estimate and the full calculation differ by less than 0.2 percent, but for other energies the difference can be larger.

## Related formulas

- [Schrödinger Equation](https://formulabro.yunusemrevurgun.com/quantum-mechanics/schrodinger-equation/)
- [Born Rule](https://formulabro.yunusemrevurgun.com/quantum-mechanics/born-rule/)
- [De Broglie Wavelength](https://formulabro.yunusemrevurgun.com/quantum-mechanics/de-broglie-wavelength/)

## Sources

- [Quantum tunnelling, Wikipedia](https://en.wikipedia.org/wiki/Quantum_tunnelling)
- [Scanning tunneling microscope, Wikipedia](https://en.wikipedia.org/wiki/Scanning_tunneling_microscope)
- [Alpha decay, Wikipedia](https://en.wikipedia.org/wiki/Alpha_decay)

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