The half life of a radioactive material is 6.93 hour. After how many hours will only one-twentieth of the material be left over? Take `log_e` (20) = 3.0.

Options

A. a. 15h
B. b. 20h
C. c. 25h
D. d. 30h

Answer

D

Explanation

For the material , `T_(1//2) = 6.93 h`
`therefore lamda = (0.693)/(6.93) = 0.1` per hour
For the decay , `N/N_0 = e^(-lamdat) therefore 1/20 = e^(-lamdat)`
`therefore e^(lamdat) = 20 therefore lamda t = log_e (20) = 3 ` (given)
`therefore 0.1xxt = 3 therefore t = (3)/(0.1) = 30` hours