Question 13 Niobium crystallises in body–centred cubic structure. If density is 8.55 g cm–3, calculate atomic radius of niobium using its atomic mass 93 u. Chapter 1: the Solid State Chemistry Class 12 solutions are developed for assisting understudies with working on their score and increase knowledge of the subjects. Niobium crystallises in body–centred cubic structure, calculate atomic radius of niobium using its atomic mass is solved by our expert teachers. You can get ncert solutions and notes for class 12 chapter 1 absolutely free. NCERT Solutions for class 12 Chemistry Chapter 1: the Solid State is very essencial for getting good marks in CBSE Board examinations
Question 13 Niobium crystallises in body–centred cubic structure. If density is 8.55 g cm–3, calculate atomic radius of niobium using its atomic mass 93 u.
Answer Edge of length of cell a = ?
Density p = 8.55 g /cm3
Number of atoms in unit cell of BCC lattice Z = 2
Avogadro number NA = 6.022x1023
Use formula
Density $p=/frac{ZM}{{{a}^{3}}{{N}_{A}}}$
Cross multiply we get
pa3NA= ZM
Divide by PNA we get
/[{{a}^{3}}=/frac{MZ}{p{{N}_{A}}}/]
Plug the values we get
/[{{a}^{3}}=/frac{93x2}{8.55x6.022x{{10}^{23}}}c{{m}^{3}}/]
/[{{a}^{3}}=3.61x{{10}^{-23}}c{{m}^{3}}/]
Take cube root both side we get
/[a={{/left( 3.61x{{10}^{-23}} /right)}^{1/3}}cm/]
/[a={{/left( 36.1x{{10}^{-24}} /right)}^{1/3}}c{{m}^{{}}}/]
/[a=/left( 36.1{{)}^{1/3}}x({{10}^{-24/3}}) /right)c{{m}^{{}}}/]
/[a=/left( 3.30x{{10}^{-8}} /right)c{{m}^{{}}}/]
For BCC
$r=/frac{a/sqrt{3}}{4}$
Plug the value of a we get
$r=/frac{3.30x{{10}^{-8}}x1.732}{4}cm$
$r=1.43x{{10}^{-8}}cm$
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