Question

If radius of the $ {}_{13}^{27}Al $ is taken to be $ {{R}_{A1}} $, then the radius of $ {}_{52}^{125}Te $ nucleus is nearly
 $ A.\dfrac{3}{5}{{R}_{A1}} $
 $ B.{{\left( \dfrac{13}{53} \right)}^{\dfrac{1}{3}}}{{R}_{A1}} $
 $ C.{{\left( \dfrac{53}{13} \right)}^{\dfrac{1}{3}}}{{R}_{A1}} $
 $ D.\dfrac{5}{3}{{R}_{A1}} $

Answer 1

Hint: This problem deals with the size of atomic radius. We have to apply the relationship of radius of atom with mass number of the atom. Mass number of atoms is defined as the sum of atomic number and number of neutrons. Atomic radius is very small as compared to the size of atoms. It is generally measured in nanometres.

Formula used:
To solve this problem we are going to use the following relation:-
 $ R={{R}_{o}}{{A}^{\dfrac{1}{3}}} $

Complete step-by-step answer:
From the above given problems we have following parameters with us:-
Atomic number of $ Al $ is $ 13 $
Atomic number of $ Te $ is $ 52 $
Mass number of $ Al $, $ {{A}_{Al}} $ is $ 27 $
Mass number of $ Te $, $ {{A}_{Te}} $ is $ 125 $
We will use the following relation:-
 $ R={{R}_{o}}{{A}^{\dfrac{1}{3}}} $
Where, $ R $ is radius of the atom, $ {{R}_{o}} $ is Fermi constant and $ A $ is mass number of the atom.
For Aluminium we have, $ {{A}_{Al}}=27 $. Radius of Aluminium $ {{R}_{Al}} $ will be given as follows:-
 $ {{R}_{Al}}={{R}_{o}}A_{Al}^{\dfrac{1}{3}} $
 $ {{R}_{Al}}={{R}_{o}}{{(27)}^{\dfrac{1}{3}}} $
 $ {{R}_{Al}}={{R}_{o}}\times 3 $ …………….. $ (i) $
For Tellurium we have, $ {{A}_{Te}}=125 $, Radius of Tellurium $ {{R}_{Te}} $ will be given as follows:-
 $ {{R}_{Te}}={{R}_{o}}A_{Te}^{\dfrac{1}{3}} $
 $ {{R}_{Te}}={{R}_{o}}{{(125)}^{\dfrac{1}{3}}} $
 $ {{R}_{Te}}={{R}_{o}}\times 5 $ ………………… $ (ii) $
On dividing $ (ii) $ by $ (i) $ we get,
 $ \dfrac{{{R}_{Te}}}{{{R}_{Al}}}=\dfrac{{{R}_{o}}\times 5}{{{R}_{o}}\times 3} $
On simplification we get,
 $ \dfrac{{{R}_{Te}}}{{{R}_{Al}}}=\dfrac{5}{3} $
 $ {{R}_{Te}}=\dfrac{5}{3}{{R}_{Al}} $
Hence, option $ (D) $ is correct.

So, the correct answer is “Option D”.

Additional Information: All of us are very familiar with the element Aluminium which is a metal of atomic number $ 13 $ and mass number $ 27 $. But very few of us would have knowledge about Tellurium. Tellurium is an element which is a metalloid having atomic number $ 52 $ and mass number $ 125 $.

Note: In solving these types of problems we should take care about atomic number and mass number of the element. Any element of the symbol $ X $ is represented as $ {}_{Z}^{A}X $ where, $ Z $ denotes atomic number of the atom and $ A $ denotes mass number of the given atom. Never be confused between $ A $ and $ Z $. It should also be noted that Fermi constant is constant for every atom.