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A proton with a kinetic energy T- 1.5MeV us captured by a deuteron H^(2). Find the excitation energy of the formed nucleus. |
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Answer» Solution :The REACTION is `p+drarrHe^(3)` Excitation energy of `He^(3)` is just the energy available in centre of mass. The velocity of the centre of mass is `(SQRT(2m_(p)T_(p)))/(m_(p)+m_(d))~~(1)/(3)sqrt((2T_(p))/(m_(p)))` In the `CM` frame, the kinetic enrgy available is `(m_(d)~~2m_(p))` `(1)/(2)m_(p)((2)/(3)sqrt((2T_(p))/(m_(p))))^(2)+(1)/(2)2m_(p)((1)/(3)sqrt((2T)/(m_(p))))^(2)=(2T)/(3)` The total energy available is then `Q=(2T)/(3)` where `Q=c^(2)(Delta_(n)+DELTA(d)-Delta_(He)^(3))` `=c^(2)xx(0.00783+0.01410-0.01603)am u` `= 5.49MeV` Finally `E=6.49MeV` |
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