Answer to Question #261434 in Inorganic Chemistry for shamsa

Question

2. Calculate:

a. The rate constant for the radioactive disintegration of cobalt-60, an isotope used in

cancer therapy.

60Co27 decays with a half-life of 5.2 years to produce

60Ni28.

b. The fraction and the percentage of a sample of the 60Co27 isotope that will remain

after 15 years.

c. How long it takes for a sample of 60Co27 to disintegrate to the extent that only 2 % of

the original amount remains?

d. The half-life of 216Po84 is 0.16 s. How long does it take to reduce a sample of it to the

negligible amount of 0.000010 % of the original amount (1.0 x 10

-7

times the original

amount)?�

Answer 1

a)Rate constant =0.13326

b)Percentage of a sample=13.553%

c)Time =29.3616yrs

d)Time=3.722secs