Question

The half-life of 9038 Sr is 28 years. What is the disintegration rate of15 mg of this isotope?

Answer 1

Given: The half-life of S 18 90 r is 28 years and the mass of isotope is 15 mg.

The half-life of a radioactive element is given as,

T 1 2 = 0.693 λ

Where, the decay constant is λ.

By substituting the value in the above equation, we get

28×365×24×60×60= 0.693 λ 8.83× 10 8 = 0.693 λ λ= 0.693 8.83× 10 8   s −1

The rate of disintegration is given by,

dN dt =λN(1)

Where, the disintegration rate is dN dt and number of initial atoms are N.

By substituting the values in the above equation, we get

dN dt = 0.693 8.83× 10 8 N(2)

Since 90 g of S 18 90 r contains 6.023× 10 23 atoms, then 15 mg of S 18 90 r contains,

N= 6.023× 10 23 ×15× 10 −3 90 =1.0038× 10 −20  atoms

By substituting the given values in equation (2), we get

dN dt = 0.693×1.0038× 10 20 8.83× 10 8 =7.878× 10 10   atoms/s

Thus, the disintegration rate of the S 18 90 r is 7.878× 10 10   atoms/s .