Question

The decay of radioactive Rn-220 to Po-216 has a half life of 55.6 seconds. If one starts with a 1.00 mg sample of Rn-220, how much is left after 10.0 minutes?

Answer 1

Substituting the given value of t_1/2 ( = 55.6 sec. )  in the equation ,

disintegration constant , = 0.693 / t_1/2

.......................................................... = 0.693 / 55.6 sec.^-1

Again,    = 2.303 / t [ log a / (a-x ) ]

where (a) represents mass of the radioactive substance at t= 0 ( initial amount ) given as = 1.00 mg.

& (a - x ) represents mass or amount of the radioactive substance at time t ( given as 10 minutes,

or , = (10 x 60 ) sec.; ( a-x ) is to be calculated.

Therefore, substituting the given values in consistent units of time and mass , we get-

.............0.693 / 55.6 = 2.303 / ( 10 x 60 ) log{( ( 1.0 /( a-x ) }

.......................( a- x ) = 5.6583 x 10^-4  mgms.

So , the amount of Rn-220 left after 10 minutes is 5.6583 x 10^-4  mgms.