Question

3 attempts left Check my work If the amount of a radioactive element decreases from 128 g to 0.50 g in 24 days, what is its half-life? days in one half-life

Answer 1

The radioactive element undergoes degradation by 1st order kinetics.

The disintegration constant for the first order kinetics for the radioactive nuclei is:

k= 2.303/t2-t1 * (log N1/N2)

or, k= 2.303/ t2-t1 * log (w1/w2)

Where, k = rate constant

t2 = final time

t1 = initial time

w1 = initial mass

w2 = final mass

Now, k= 0.693/t1/2 for the first order reaction where t1/2 is the half life.

So,

0.693/t1/2 = 2.303/t2-t1 * log(w1/w2)

Putting all the values, we get

0.693/t1/2 = 2.303/24 * log(128/0.50)

=> 0.693/t1/2 = 0.095958 * log256

=> 0.693/t1/2 = 0.23109

=> t1/2 = 0.693/0.23109

=> t1/2 = 3 days.

from first order rabe constant formula, K = N 2.303 (t₂-t) 0.693 = 2.303 log Wie ti/2 lt₂-ti) o wz Putting the values , we get 0.69 = 2.303 log 1200 +1/2 24 log 12150 24 0.693 = 0.095958 X log 256 til2 = 0.693 +1/2 0.93109 t1/2 = 0.693 0.23109 days [tila = 3 days