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The half-life a radioacitve substance is `40` yeard. How long will it take to reduce to one fourth of its original amount and what is the value of decay constant ?A. `40 yr,0.9173//yr`B. `90yr,9.017//yr`C. `80yr,0.0173//yr`D. None of these |
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Answer» Correct Answer - C To reduce one-fourth it takes time, `t=2(T_(1//2))=2xx40=80yr` So, decay constant, `lamda(0.693)/(T_(1//2))=(0.963)/(40)=0.0173yr` |
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