Question

26. The half-life of a radioisotope is 10 h. Find the total
number of disintegrations in the tenth hour measured
from a time when the activity was 1 Ci.​

Answer 1

Answer:

Half-life of radioisotopes, T1/2=10hrs

Half-life of radioisotopes, T1/2=10hrsInitial activity, A0=1Ci

Half-life of radioisotopes, T1/2=10hrsInitial activity, A0=1CiDisintegration constant, λ=10×36000.693s−1

Half-life of radioisotopes, T1/2=10hrsInitial activity, A0=1CiDisintegration constant, λ=10×36000.693s−1After 9 hours 

Half-life of radioisotopes, T1/2=10hrsInitial activity, A0=1CiDisintegration constant, λ=10×36000.693s−1After 9 hours Activity of radioactive sample,

Half-life of radioisotopes, T1/2=10hrsInitial activity, A0=1CiDisintegration constant, λ=10×36000.693s−1After 9 hours Activity of radioactive sample,Activity, A=A0e−λt

Half-life of radioisotopes, T1/2=10hrsInitial activity, A0=1CiDisintegration constant, λ=10×36000.693s−1After 9 hours Activity of radioactive sample,Activity, A=A0e−λt=1×e−10×36000.693×9=0.536Ci

Half-life of radioisotopes, T1/2=10hrsInitial activity, A0=1CiDisintegration constant, λ=10×36000.693s−1After 9 hours Activity of radioactive sample,Activity, A=A0e−λt=1×e−10×36000.693×9=0.536Ci∴ Number of atoms left, N=λA=0.693=103.0230.536×10×3.7×1010×3600 After 10 hrs

Half-life of radioisotopes, T1/2=10hrsInitial activity, A0=1CiDisintegration constant, λ=10×36000.693s−1After 9 hours Activity of radioactive sample,Activity, A=A0e−λt=1×e−10×36000.693×9=0.536Ci∴ Number of atoms left, N=λA=0.693=103.0230.536×10×3.7×1010×3600 After 10 hrsActivity, A"=A0e−λt=1×e−100.693×10=0.5Ci

Half-life of radioisotopes, T1/2=10hrsInitial activity, A0=1CiDisintegration constant, λ=10×36000.693s−1After 9 hours Activity of radioactive sample,Activity, A=A0e−λt=1×e−10×36000.693×9=0.536Ci∴ Number of atoms left, N=λA=0.693=103.0230.536×10×3.7×1010×3600 After 10 hrsActivity, A"=A0e−λt=1×e−100.693×10=0.5CiNumber of atoms left after the 10th hour (N") will be 

Half-life of radioisotopes, T1/2=10hrsInitial activity, A0=1CiDisintegration constant, λ=10×36000.693s−1After 9 hours Activity of radioactive sample,Activity, A=A0e−λt=1×e−10×36000.693×9=0.536Ci∴ Number of atoms left, N=λA=0.693=103.0230.536×10×3.7×1010×3600 After 10 hrsActivity, A"=A0e−λt=1×e−100.693×10=0.5CiNumber of atoms left after the 10th hour (N") will be A"=λN"

Half-life of radioisotopes, T1/2=10hrsInitial activity, A0=1CiDisintegration constant, λ=10×36000.693s−1After 9 hours Activity of radioactive sample,Activity, A=A0e−λt=1×e−10×36000.693×9=0.536Ci∴ Number of atoms left, N=λA=0.693=103.0230.536×10×3.7×1010×3600 After 10 hrsActivity, A"=A0e−λt=1×e−100.693×10=0.5CiNumber of atoms left after the 10th hour (N") will be A"=λN"N"=λA"

Half-life of radioisotopes, T1/2=10hrsInitial activity, A0=1CiDisintegration constant, λ=10×36000.693s−1After 9 hours Activity of radioactive sample,Activity, A=A0e−λt=1×e−10×36000.693×9=0.536Ci∴ Number of atoms left, N=λA=0.693=103.0230.536×10×3.7×1010×3600 After 10 hrsActivity, A"=A0e−λt=1×e−100.693×10=0.5CiNumber of atoms left after the 10th hour (N") will be A"=λN"N"=λA"=0.693/100.5×3.7×1010×3.600=26.37×