Question

The radioactive decay of radium ²²⁶Ra follows a first-order rate law.

This decay process has a half-life of 1600 years.

(a). What is the value of the rate constant (expressed in s^-1) for this decay process?

(b). How many disintegrations per second would occur in a one-gram sample of ²²⁶Ra ?

Answer 1

a)

Given:

Half life = 1600 year

1 year = 3.154*10^7 s

so,

half life = 1600 * 3.154*10^7 s = 5.0464*10^10 s

use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(5.0464*10^10)

= 1.373*10^-11 s-1

Answer: 1.373*10^-11 s-1

b)

number of atoms = number of mol * Avogadro’s number

number of atoms = (mass / molar mass) * 6.022*10^23 atoms

number of atoms = (1 / 226) * 6.022*10^23 atoms

number of atoms = 2.66*10^21 atoms

disintegration per second = number of atoms * rate constant

= 2.66*10^21 atoms * 1.373*10^-11 s-1

= 3.66*10^10 disintegration per second

Answer: 3.66*10^10 disintegration per second