Home | | Physics 12th Std | Radioactivity: Alpha decay

## Chapter: 12th Physics : UNIT 9 : Atomic and Nuclear Physics

When unstable nuclei decay by emitting an α-particle, it loses two protons and two neutrons.

Alpha decay

When unstable nuclei decay by emitting an α-particle ( 42He nucleus), it loses two protons and two neutrons. As a result, its atomic number Z decreases by 2, the mass number decreases by 4. We write the alpha decay process symbolically in the following way Here X is called the parent nucleus and Y is called the daughter nucleus.

Example: Decay of Uranium 23892U to thorium 23490Th with the emission of 42He nucleus (α-particle)

23892U 23490Th + 42He

As already mentioned, the total mass of the daughter nucleus and 42He nucleus is always  less than that of the parent nucleus. The difference in mass ( ∆m = mX mY mα ) is released as energy called disintegration energy Q and is given by

Q = (mX mY mα )c2           (8.27) Note that for spontaneous decay (natural radioactivity) Q >0. In alpha decay process, the disintegration energy is certainly positive (Q > 0). In fact, the disintegration energy Q is also the net kinetic energy gained in the decay process or if the parent nucleus is at rest, Q is the total kinetic energy of daughter nucleus and the 42He nucleus. Suppose Q < 0, then the decay process cannot occur spontaneously and energy must be supplied to induce the decay.

In alpha decay, why does the unstable nucleus emit 42He nucleus? Why it does not emit four separate nucleons? After all He consists of two protons and two neutrons. For example, if 92238U nucleus decays into 90234Th by emitting four separate nucleons (two protons and two neutrons), then the disintegration energy Q for this process turns out to be negative. It implies that the total mass of products is greater than that of parent( 92238U ) nucleus.

This kind of process cannot occur in nature because it would violate conservation of energy. In any decay process, the conservation of energy, conservation of linear momentum and conservation of angular momentum must be obeyed.

EXAMPLE 8.11

(a) Calculate the disintegration energy when stationary 23292U nucleus decays to thorium 22890Th with the emission of α particle. The atomic masses are of 23292U = 232.037156 u , 22890Th = 228.028741u and 42He = 4.002603u

(b) Calculate kinetic energies of 22890Th and α-particle and their ratio.

Solution

The difference in masses

m = (mU mTh - mα)

= (232.037156–228.028741 – 4.002603)u

The mass lost in this decay = 0.005812 u

Since 1u = 931MeV, the energy Q released is

Q = (0.005812u)×(931MeV / u)

= 5.41MeV

This disintegration energy Q appears as the kinetic energy of α particle and the daughter nucleus.

In any decay, the total linear momentum must be conserved.

Total linear momentum of the parent nucleus = total linear momentum of the daughter nucleus +α particle

Since before decay, the uranium nucleus is at rest, its momentum is zero.

By applying conservation of momentum, we get It implies that the alpha particle and daughter nucleus move in opposite directions.

In magnitude mα υα = mThυTh

The velocity α particle υα =  [mTh/mα] υTh.

Note that mTh /ma > 1, so υα > υTh . The ratio of the kinetic energy of α particle to the daughter nucleus By substituting, the value of υα into the above equation, we get The kinetic energy of α particle is 57 times greater than the kinetic energy of the daughter nucleus (22890Th ).

The disintegration energy Q = total kinetic energy of products

K.Eα + K.ETh = 5.41MeV

57K.ETh + K.ETh = 5.41MeV

K.ETh = 5.41/58 MeV = 0.093MeV

K.Eα = 57K.ETh = 57×0.093 = 5.301MeV

In fact, 98% of total kinetic energy is taken by the α particle.

Tags : Nuclear Physics , 12th Physics : UNIT 9 : Atomic and Nuclear Physics
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
12th Physics : UNIT 9 : Atomic and Nuclear Physics : Radioactivity: Alpha decay | Nuclear Physics