Equal amounts of two samples of carbon were burnt and the radioactivities ofCO_2 formed were measured. The activitied were 2100and 1400 counts per week respectively.What will be the age difference of the sample. [ t_(1//2) forC^(14) = 5600 years ]

Equal amounts of two samples of carbon were burnt and the radioactivit

1.	Equal amounts of two samples of carbon were burnt and the radioactivities ofCO_2 formed were measured. The activitied were 2100and 1400 counts per week respectively.What will be the age difference of the sample. [ t_(1//2) forC^(14) = 5600 years ]
Answer» 2002 years 2345 years 4250 years 3343 years Solution :`t_(1//2) = 5600` years Decay constant , `LAMBDA = (0.693)/(t_(1//2)) = (0.693)/(5600) = 1.24 xx 10^(-4) "year"^(-1)` For first sample : `(N_0)/(N) = ("NUMBER of counts per week of CARBON INITIALLY")/("number of counts per week of carbon finally")` `= (N_0)/(2100) "".......(i)` For second sample : `(N_0)/(N) = (N_0)/(1400) "" .......(ii)` The AGE difference of the sample : `t = (2.303)/(lambda) ["log" (N_0)/(1400) - "log" (N_0)/(2100)]` `= (2.303)/(1.24 xx 10^(-4))["log" N_0 - "log" 1400 - "log" N_0 + log 2100]` `= (2.303)/(1.24 xx 10^(-4)) [log 2100 - log 1400]` `= (2.303)/(1.24 xx 10^(-4)) [3.32 - 3.14]` `= t = (2.303)/(1.24 xx 10^(-4)) xx 0.18 = 3343.06` years