Question

Only lanthanide which is radioactive is:
A) $\text{ Sm }$
B) $\text{ Yb}$
C) $\text{ Pm }$
D) \[\text{ Eu}\]

Answer 1

Hint: The lanthanides are the elements in which the last electron enters the f-orbital. The lanthanides are not radioactive except the one. The element exhibits radioactivity due to the presence of several protons and neutrons.

Complete step by step answer:
Lanthanides are also known as the rare earth metal. The lanthanides are the elements with the atomic number from 58 to 71 which is from Lanthanum $\text{ (La) }$ to Lutetium$\text{ (Lu) }$. In the lanthanide series, the valence shell electron is filled in the $\text{ 4f }$ energy level.
The lanthanides have the general valence shell configuration as:\[\text{ }\left[ \text{Xe} \right]\text{ 4f}{{\text{ }}^{\text{n}}}\text{ 6}{{\text{s}}^{\text{2}}}\] .
The promethium $\text{ Pm }$ has an atomic number of 61. The electronic configuration of promethium is as follows:
\[\text{Pm = }\left[ \text{Xe} \right]\text{ 4}{{\text{f}}^{\text{5}}}\text{ 6}{{\text{s}}^{\text{2}}}\]
Promethium $\text{ (Pm) }$ is the only lanthanide that exhibits radioactivity. It is one of the two elements among the first 83 elements. It does not have a stable or long-lived isotope.
The promethium isotope (promethium-146) is unstable. The primary decay product of promethium is neodymium $\text{ (Nd) }$ and samarium$\text{ (Sm) }$.
The promethium -146 decay into the lighter neodymium via a positron decay and the electron capture results in the formation of a heavier isotope to samarium via beta decay. The reactions are as shown below,
$_{61}^{149}\text{ Pm}\xrightarrow{{{\beta }^{+}}}\text{ }_{60}^{149}\text{ Nd }$
Positron decay is associated with the removal of an electron from the isotope. Here, the promethium loses an electron and forms a lighter isotope.
$\text{ }_{61}^{149}\text{ Pm}\xrightarrow{{{\beta }^{-}},\text{ 53}\text{.1 hr}}\text{ }_{62}^{149}\text{ Sm }$
The promethium isotope, the isotope captures electrons and forms heavier isotope samarium.
Promethium may decay to the other promethium isotope. The one promethium isotope can undergo the alpha decay to form a stable praseodymium-141.
The promethium has a stable isotope is$\text{ Pm-145 }$. It has a half-life of $\text{ 17}\text{.7 }$ years. It has 84 neutrons and it emits the alpha particle to form the praseodymium -141 .the reaction is as follows;
 $\text{ }_{61}^{145}\text{ Pm}\xrightarrow{\alpha -decay\text{ , }}\text{ }_{59}^{141}\text{ Sm + }_{2}^{4}\text{He}$
The promethium has the 18 nuclear isomers with the mass number of 133 to 142, 144, 148, 152, and 154. The most stable isotope is promethium-148 which has a half-life of $43.1$ days. Thus, from given lanthanides, only promethium $\text{ (Pm) }$can undergo the decay and it is radioactive.

Hence, (C) is the correct option.

Note: The promethium is the only self-made radioactive lanthanide. Except the promethium, all other lanthanides are non-radioactive. However, on the other hand, the actinides are highly radioactive. This is because the actinides have a higher charge and smaller size (due to actinide contraction) of their atoms.