Question

(a) 1. Which element is used in the lining of the special aprons worn by workers in nuclear power plants?
2. Why is this element preferred?
(b) \(^{24}_{11}\)Na emits a nuclear radiation which does not alter the mass number and is deflected by a magnetic field.
1. Name the type of nuclear radiation emitted by \(^{24}_{11}\)Na.
2. Write the equation for this radioactive decay.

Hint:

Remember the changes for nuclear decays: Alpha decay (\(A-4, Z-2\)), Beta decay (\(A\) unchanged, \(Z+1\)), Gamma decay (\(A, Z\) unchanged).

Answer 1

(a) 1. Element for shielding aprons:
The element used for lining radiation-shielding aprons is Lead (Pb).
(a) 2. Reason for preference:
Lead is preferred for radiation shielding for two main reasons: - High Density and High Atomic Number: Lead has a very high density and a large number of electrons per atom. This dense structure is extremely effective at absorbing the energy of high-energy photons like gamma rays and X-rays, significantly reducing their intensity. - Cost-Effectiveness and Malleability: It is relatively inexpensive, abundant, and easy to work with, making it practical for constructing shielding materials.
(b) 1. Type of nuclear radiation:
We are given two clues:
1. The radiation does not alter the mass number (A).
2. The radiation is deflected by a magnetic field, which means it consists of charged particles.
Let's analyze the common types of radiation:
- Alpha (\(\alpha\)) decay: Emits a helium nucleus (\(^{4}_{2}\text{He}\)). It is charged but changes the mass number by -4. So, it's not alpha.
- Gamma (\(\gamma\)) decay: Emits a high-energy photon. It does not change the mass number but is not charged, so it is not deflected. So, it's not gamma.
- Beta (\(\beta\)) decay: Emits an electron (\(^{0}_{-1}\beta\)). An electron is a charged particle, so it will be deflected. Its mass number is 0, so emitting it does not alter the mass number of the nucleus. This matches both conditions. Therefore, the radiation is a Beta particle.
(b) 2. Equation for the decay:
In beta decay, a neutron within the nucleus converts into a proton and an electron. The electron is ejected as a beta particle.
- The mass number (A) remains unchanged: \(A = 24\).
- The atomic number (Z) increases by 1: \(Z_{new} = 11 + 1 = 12\).
- The element with atomic number 12 is Magnesium (Mg).
The balanced nuclear equation is:
\[ ^{24}_{11}\text{Na} \rightarrow ^{24}_{12}\text{Mg} + ^{0}_{-1}\beta + \bar{\nu}_e \] (where \(\bar{\nu}_e\) is an electron antineutrino, often omitted in introductory-level equations). The required equation is: \(^{24}_{11}\text{Na} \rightarrow ^{24}_{12}\text{Mg} + ^{0}_{-1}\beta\).