Question

Question 2.37 The diameter of zinc atom is.Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

Class XI Structure of Atom Page 67

Answer 1

Here ,
Diameter of Zn - atom = 2.6 A° = 2.6 × 10^-10 m
radius of Zn - atom = 2.6/2 × 10^-10 m
radius of Zn - atom = 1.3 × 10^-10 m

A/C to question,
number of atoms present in a length = 1.6 cm
= 1.6 × 10^-2 m

$number\: of\: Zn - atoms=\frac{total \: length}{diameter \: of \: atom}$

= 1.6 × 10^-2 m/2.6 × 10^-10 m
= 0.6154 × 10^8 atoms
= 6.154 × 10^7 atoms

Answer 2

Answer: (A)1.3*10^2 pm

     (B)      6.15 *10^7 atoms

Explanation: hey!!!!!

(A) to find : radius of zinc atom in pm

   given that : diameter= 2.6 A i.e 2.6 *10^-10m

radius =diameter /2

     2.6*10^-10/2 = 1.3*10^-10 m

in pm: as we know ; 1 m =1/10^-12pm

so; 1.3*10^-10/10^-12

   answer; 1.3*10^2pm

(B)  to find; no.of zinc atoms

given that:  length =1.6 cm

n*d = length

n*2.6*10^-8cm= 1.6

n = 0.6 *10^8 atoms

         or

6.15 *10^7 atoms

    hope it helps.........