Question

What is meant by the Alpha decay? Estimate the velocity or energy of α-particle.

Answer 1

Alpha decay

Alpha decay is a process involving the emission of a fast-moving helium nucleus (an alpha-particle) by nuclei which generally contain 210 or more nucleons.

Since ₂He⁴ contains two protons and two neutrons, after an alpha emission the parent nucleus is transformed into a daughter nucleus which has (i) an atomic number smaller by two and (ii) mass number smaller by four.

Transformation of the _ZX^A nucleus into the _Z-2X^A-4 nucleus by an alpha decay can be expressed by the equation.

_ZX^A = _Z-2Y^A-4 + ₂He⁴ ..........(1)

Energy released. The energy Q, released in this process, can be obtained from Einstein's mass energy relation.

It is given by the expression

Q = (M_X - M_Y - M_α) c²

This energy is shared by the daughter nucleus, _Z-2Y^A-4 and the alpha-paraticle, ₂He⁴.

As the parent nucleus _ZX^A is at rest before it undergoes alpha-decay, alpha-particles are emitted with fixed energy, which can be calculated by applying the principle of conservation of energy and momentum.

Let v_α and v_y be the velocities of the alpha-particles and the daughter nucleus, _Z-2Y^A-4. According to the principle of conservation of momentum,

M_Y. v_Y = M_α. v_α ...........(2)

By equating the sum of kinetic energies of the nucleus Y and the alpha particle of the released in the alpha-decay, we have another equation.

\(\frac{1}{2}\) M_α.v_α² + \(\frac{1}{2}\) M_Yv² = Q ...........(3)

By substituting for vy from Eq. (2) in Eq. (3), we can easily obtain

If we substitute M_y = A - 4 amu and M_α = 4 amu in Eq. (4) the kinetic energy carried by the alpha-particle can be approximated by the relation.

[K.E.]α = Q \(\frac{(A-4)}{A}\) ............(5)