Physics [5 Mark Question Answer]

$\ce{\underset{\text{(helium isotope)}}{^3_2He} + \underset{\text{(deuterium)}}{^2_1H} -> \underset{\text{(helium) }}{^4_2He}} + \underset{\text{(proton)}}{^1_1H} + 18.3 \text{MeV}$

In all three deuterium nuclei fuse to form a helium nucleus with a release of 21·6 MeV energy.

When two deuterium nuclei $\ce {^2_1H}$ fuse , nucleus of helium isotope $\ce {^3_2He}$ is formed and 3·3 MeV energy is released. This helium isotope again gets fused with one deuterium nucleus to form a helium nucleus $\ce {^4_2He}$ and 18·3 MeV of energy is released in this process.

Nuclear fission is the process in which a heavy nucleus is splits into two light nuclei nearly of the same size by bombarding it with slow neutrons.

When uranium with Z = 92 is bombarded with neutron, it splits into two fragments namely barium (Z = 56) and krypton (Z = 36) and a large amount of energy is released which appears due to decrease in the mass

$\ce {^235_92 U + ^1_0n -> (^236_92U) -> ^144_56Ba + ^89_36Kr + 3^1_0n + Energy}$

The following changes occur when an atom emits

An alpha particle: atomic number decreases by 2 and mass number decreases by 4.

Example: $\ce{^238_92 U -> ^234_90 Th + ^4_2 He}$

A beta particle: atomic number increases by one, but mass number does not change.

Example: $\ce{^14_6 C -> ^14_7 N + ^0_-1 β}$

Gamma particle: it does not change anything in the nucleus, the energy of the nucleus decreases.

number does not change.

Example: $\ce{^A_Z X^* -> ^A_Z X + γ}$

After emitting an alpha particle the daughter element occupies two places to the left of the parent element in the periodic table.

Reason: If a parent nucleus X becomes a new daughter nucleus Y as a result of -decay, then the -decay can be represented as:

$\ce{\underset{\text{parent nucleus}}{^A_Z X} ->[Alpha decay] \underset{\text{Daughter nucleus}}{^{A-4}_{z-2}Y} + \underset{\text{α-particle}}{^4_2 He}}$

Thus, the resulting nucleus has an atomic number equal to (Z-2). Hence, it shifts two places to the left of the parent element in the periodic table.

(b) After emitting a β

-particle, the daughter element occupies one place to the right of the parent element in the periodic table.

Reason: If a parent nucleus X becomes a new daughter nucleus Y as a result of -decay, then the -decay can be represented as:

$\ce{\underset{\text{parent nucleus}}{^A_Z X} ->[β particle] \underset{\text{Daughter nucleus}}{^{A}_{z+1}Y} + \underset{\text{β-particle}}{^0_-1e}}$

Thus, the resulting nucleus has an atomic number equal to (Z+1). Hence, it shifts one place to the right of the parent element in the periodic table.

(c) After emitting -radiation, the element occupies the same position in the periodic table.

Reason: If a parent nucleus X becomes a new daughter nucleus Y as a result of -decay, then the -decay can be represented as:

$\ce{\underset{\text{parent nucleus}}{^A_Z X} ->[γ particle] \underset{\text{Daughter nucleus}}{^{A}_{z}Y} + \underset{\text{γ-particle}}{γ}}$

Thus, the resulting nucleus has atomic number equal to Z. Hence, it occupies the same position as the parent element in the periodic table.

If a radioactive nucleus P with mass number A and atomic number Z emits a beta particle to form a daughter nucleus Q with mass number A and atomic number Z+1, then the change can be represented as follows:

$\ce{^A_Z P ->[β particle]} $ $\ce {^A_{z+1} Q} + {^0_{-1} e} $

(a) Atomic number 'Z' is not conserved. It is increased by 1.

(b) Mass number A is conserved.