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The minimum (threshold ) KE of the proton to initiate the nulear reaction `" " p+^(7)Lirarr ""^(7)Be+n` Given `m_(p)=1.0073` amu,`m_(1) =7.0144` amu, `m_(Be)=7.0147` amu, `m_(0)=1.0087` amu.A. `2xx10^(-15) J`B. `4xx10^(-14)J`C. `2.5xx10^(-13)J`D. `8xx10^(-6)J` |
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Answer» Correct Answer - C Required energy `=931.5xx10^(6)xx1.6xx10^(-19)` `[m_(Be)+m_(a)-m_(p)-m_(Li)]=2.5xx10^(-13)` Joule |
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