1.

The minimum (threshold ) KE of the proton to initiate the nulear reaction "" p+^(7)Lirarr ""^(7)Be+n Givenm_(p)=1.0073 amu,m_(1) =7.0144 amu, m_(Be)=7.0147 amu, m_(0)=1.0087 amu.

Answer»

`2xx10^(-15) J`
`4XX10^(-14)J`
`2.5xx10^(-13)J`
`8xx10^(-6)J`

SOLUTION : Required energy `=931.5xx10^(6)xx1.6xx10^(-19)`
`[m_(Be)+m_(a)-m_(p)-m_(Li)]=2.5xx10^(-13)` JOULE


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