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A radioactive material has mean lives of 1620 yr and 520 yr for `alphaandbeta-"emission"`, respectively. The material decays by simultaneous `alphaandbeta-"emissions`. The time in which 1/4th of the material remains intact isA. 4.675 yrB. 7.20 yrC. 5.45 yrD. 324 yr |
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Answer» Correct Answer - C `lamda=(lamda_(1)+lamda_(2))=(1)/(1620)+(1)/(520)=2.54xx10^(-3)yr^(-1)` `therefore" " t_(1//2)=(1n(2))/(lamda)=272.8 yr` (1/4) th of the material remainsintact after 2 half-lives. |
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