A stationary thorium nucleus (A=220, Z=90) emits an alpha particle with kinetic energy E_{\alpha} . What is the kinetic energy of the recoiling nucleus?

^{220} Z _{90} \rightarrow \alpha + X m_X = 220 - m_{\alpha} = 216 E =\cfrac{p^2}{2m} p = \sqrt{2mE} Momentum Conservation: \sqrt{2m_{\alpha}E_{\alpha}}= \sqrt{2m_{X}E_{X}} m_{\alpha}E_{\alpha}= m_{X}E_{X} 4E_{\alpha}= 216E_{X} E(X) = \cfrac{E_{\alpha}}{54}

Nuclei - Tough concepts made easy

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Nuclei

- A deep dive into the toughest topics

Binding Energy Explained

The nucleons are held together by nuclear force, which is short ranged but much stronger than coulomb force. It is found that the mass of a nucleus is less than the sum of the mass of the individual nucleons. What does this mass defect signify?

The mass-defect is responsible for the binding together of the nucleons.

Let us understand how this mass defect arises.

When individual nucleons come together to form a nucleus, they liberate some energy in the form of photons. The liberated energy appears as a reduction in the net mass of the nucleus following Einstein's mass energy relation(

Δ E = m c^{2}

)

The same energy needs to be supplied back to the nucleus if we need to separate the nucleus into its individual nucleons. To overcome the nuclear forces holding the nucleus together, energy equivalent to the binding energy needs to be supplied.

To take as an example, let us find out the binding energy of a Helium nucleus

Energetics of Alpha Decay

Alpha Decay is the process by which an unstable nucleus decays into a daughter nucleus by emitting an alpha particle. The general form of alpha decay is

_{Z}^{A} P \to_{Z - 2}^{A - 4} D +_{2}^{4} H e

The decay of U-238 to Thorium-234 is an example of alpha decay

_{92}^{238} U \to_{90}^{234} T h +_{2}^{4} H e

The process of alpha decay is a spontaneous decay process which means no energy needs to be provided to the parent nucleus to initiate the decay.

Δ m

is the difference in mass between initial and final products.

mass defect=

Δ m = m_{P} - m_{D} - m_{α}

For alpha decay the final products have less total mass than the parent nucleus.

Δ m

is positive, and the reaction releases energy. The energy released is given by the Q-value or the reaction energy.

Q = Δ m c^{2} = (m_{P} - m_{D} - m_{α}) c^{2} > 0

As, alpha decay is spontaneous decay, the energy released appears as kinetic energy in the alpha particle and the daughter nucleus. The Q value is also the net kinetic energy gained in the decay process.

To study the energetics of alpha decay, let us apply the law of conservation of momentum.

If the parent nucleus is at rest then we can say the net momentum after decay is also zero.

p_{D} + p_{α} = 0

.............(i)

M_{D} v_{D} = - M_{α} v_{α}

It implies that the alpha particle and daughter nucleus move in opposite directions.

Conservation of energy is also applicable to any decay process.

As the initial kinetic energy is zero the Q value appears as kinetic energy of alpha particle and daughter nucleus.

Q = K_{D} + K_{α}

.............(ii)

where,

K_{D} = \frac{p _{D}^{2}}{2 M _{D}}

and

K_{α} = \frac{p _{α}^{2}}{2 M _{α}}

Substituting the values of momentum in the conservation of momentum equation we get

2 M_{D} K_{D} - 2 M_{α} K_{α}

Rearranging we get,

K_{D} = \frac{m _{α}}{m _{D}} K_{α}

substituting the value of

K_{D}

in (ii)

K_{D} = \frac{m _{α}}{m _{D}} K_{α}

Let us practice a question based on alpha decay energetics.

A stationary thorium nucleus

(A = 220, Z = 90)

emits an alpha particle with kinetic energy

E_{α}

. What is the kinetic energy of the recoiling nucleus?

\frac{E _{α}}{1 0 8}

\frac{E _{α}}{1 1 0}

\frac{E _{α}}{5 5}

\frac{E _{α}}{5 4}

View solution

Nuclear Fission

Heavy nuclei with mass numbers more than 170 have binding energies less than 8MeV. We know greater the binding energy more stable is the nucleus. Hence heavier nuclei tend to undergo nuclear reactions such that the products have higher binding energy. As the binding energy increases, there is a net release of energy in the process.

The process of breaking up of the nucleus of a heavier atom into two smaller nuclei with the release of a large amount of energy is called nuclear fission.

An example of fission reaction is splitting of U-235 when excited by a neutron.

_{92}^{235} U +_{0}^{1} n \to_{92}^{236} U^{*} \to_{56}^{141} B a +_{36}^{92} K r + 3_{0}^{1} n + Q

The following steps lead to the fission of U-235:

Neutron Capture: A slow neutron is absorbed by the the U-235 nucleus. As a result the mass number of Uranium increases by 1 and it moves into excited state.
The excited state of Uranium, $_{92}^{236} U^{*}$ lasts of about $1 0^{- 12} s$ . The exited state of Uranium is highly unstable.
Once the nucleus is excited, Uranium splits into Barium and Krypton along with the release of 3 neutrons.

Let us determine the energy released in the reaction

m (_{92}^{235} U) = 235.045733 u

m (_{0}^{1} n) = 1.008665 u

Total mass of reactant = 236.054398 u

m (_{56}^{141} B a) = 140.9177 u

m (_{36}^{91} K r) = 91.8854 u

Mass of 3 neutrons = 3.025995 u

The total mass of products = 235.829095 u

Mass Defect=

Δ m = 236.054398 u - 235.829095 u = 0.225303 u

Energy released=

0.225303 u \times 931 M e V \approx 200 M e V

The energy released appears as kinetic energy in the daughter nuclei and the neutrons. The high neutrons can further transfer energy to nearby U-235 nuclei and cause further fission reactions, which is precisely what we call chain reactions.