If the activity of Ag^108 is 3 micro curie, the number of atoms present in it are

Q. If the activity of ¹⁰⁸Ag is 3 micro curie, the number of atoms present in it are (λ = 0.005 sec^-1)

(a) 2.2 × 10⁷

(b) 2.2 × 10⁶

(c) 2.2 × 10⁵

(d) 2.2 × 10⁴

Ans: (a)

Sol: 1 Curie = 3.7×10¹⁰ dps

1 micro Curie = 3.7×10⁴ dps

Since , A = λ N

3 ×3.7×10⁴ = 5×10^-3 N

N = (3 ×3.7×10⁴)/(5×10^-3)

= 2.2 × 10⁷