# Transmission coefficient for α particle emitted from the nucleus

A radio isotope may decay into an α particle and a relaxed nucleus. Here we predict tunneling of an α particles through a coulomb potential barrier assuming free particle of energy E inside the nucleus. The alpha particle experiences the coulomb field when it comes out of the nucleus. Thus α particles behaves as entity in the nucleus.

Let α particle collide with the walls and in order to escape from the nucleus. The collision frequency and the transmission coefficient are represented by 𝛾 and T. The probability of α decay per unit time is analogous to the disintegration constant 𝜆,

𝜆 = 𝛾T

here 𝛾 = velocity of α particle / diameter of the nuclei = 𝛾 / 2 r_o .....i

The potential energy is characterized by

V(x) = 2ze² / r ; x > r_o

= constant ; x < r_o

Here z and r represent the atomic number of the relaxed nucleus and the distance from the centre of the nucleus.

Now we discuss the case E< V (i.e tunneling phenomena). The schrodinger wave equations for the region I, II and III are

Here V = constant = zero (say) for region I

= 0; nuclear force vanishes in the region III

Ko² = 2mE/ ℏ²

K² = 2m(V - E)\ℏ² ....v

The wave functions of above equations are

Here we have assume that reflection takes place at both the surface of barrier. Thus no alpha particle is coming from right. The boundary conditions are

Applying these boundary conditions on equation vi, vii, viii we get

Solving these equations for A and A' we get

Now the transmission coefficient,

T = l A'/ Al ²

= 4E (V - E) / 4E(V - E) + V²sin²h KL ....xvi

We have

sin hkL = e^kL + e^-kL / 2 -> 0 as kL -> ∞

when kL >> 1

sin hkL = e^kL / 2

Thus 1/ sinh²kL = { 1 / e^kL/2}² = 4e^-2kL

Then equation xvi becomes

Taking log on both sides

ln T = ln [ 16 E(V- E)/ V² ] - 2kL which is smaller than kL, E< V and kL>>1....xvii

In case of α particle, equation xvii can be written as

Here m = mass of the α particle

E = energy of the α particle

V = Binding energy of the nuclei

When r = r_1, E= V

E = 2ze²/ r_1

and V = 2ze²/r .....xix

Substituting the value of E and V from eq xix

Put r/r_1 = cos 2𝜃

dr = -r_1 2 cos 𝜃 sin 𝜃 d𝜃 = - r_1sin 2𝜃 d𝜃

At r= r_1, cos 𝜃 = 1 => 𝜃 = 0°

Substituting these values in equation xx

The width of the potential barrier is extremely large compared to the radius of the nuclei ie r >> r_0

Equation xxi becomes

Equation xxii represents the empirical formula for the transmission coefficient of α decay. This expression is known as Geiger- Nuttal law.

Understanding the Geiger-Nuttall law helps in studying the energetics and probabilities associated with alpha decay processes, which are important in fields such as nuclear physics, radiochemistry, and radiobiology.

This note is a part of the Physics Repository.