Question

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`

Answer 1

Correct Answer - C
Required energy `=931.5xx10^(6)xx1.6xx10^(-19)`
`[m_(Be)+m_(a)-m_(p)-m_(Li)]=2.5xx10^(-13)` Joule