20 mg of C-14 has half-life of 5760 yr. 100 mg of sample containing C-14 is reduced to 25 mg in

Options

A. 5760 yr
B. 11520 yr
C. 17280 yr
D. 23040 yr

Answer

C

Explanation

Half life of radioactive
`t_(1//2)=5760` yr,`[R]_(0)` =25 mg
Rate constant (k)=`(0.6932)/(t_(1//2))=(06932)/(5760)"yr"^(-1)`
`k=(2.303)/(t)log[R]/([R])`
`(R_(0)=200` mg initial amount)
`(0.6932)/(5760)=(2.303)/(t)"log"(200)/(25)`
(R=25 mg reduced amount)
`=(2.303)/(t)log 8`
`(0.693)/(5720)=(2.303)/(t)xx0.9030`
`t=(5760xx2.303xx0.9030)/(0.6932)`
`(11978.54)/(0.6932)`=17280 yr`