A neutron is absorbed by a `._3Li^6` nucleus with subsequent emission of an alpha particle. Write the corresponding nuclear reaction. Calculate the energy released in this reaction.
`m(._3Li^6)=6.015126u, m(._2He^4)=4.0026044u`
`m(._0n^1)=1.0086654u, m(._1He^3)=3.016049u`
Take `1u=931MeV`.

A neutron is absorbed by a ._(3)^(6)Li nucleus with the subsequent emission of an alpha particle. (i) Write the corresponding nuclear reaction. (ii) Calculate the energy released, in MeV , in this reaction. [Given: mass ._(3)^(6)Li=6.015126u , mass (neutron) =1.0086654 u mass (alpha particle) =0.0026044 u and mass (triton) =3.010000 u . Take i u=931 MeV//c^(2) ]

Calculate the energy released in the following reaction ._3Li^6 + ._0n^1 to ._2He^4 + ._1H^3

Calculate the energy equivalent of 1 a.m.u. in MeV

In a star, three alpha particle join in a single reaction to form ._(6)C^(12) nucleus. Calcuate the energy released in the reaction Given : m(._(2)He^(4)) = 4.002604 am u, " " m(._(6)C^(12)) = 12.000000 am u

A slow neutron strikes a nucleus of ._(92)U^(235) splitting it into lighter nuclei of barium and krypton and releasing three neutrons. Write the corresponding nuclear reaction. Also calculate the energy released on this reaction Given m(._(92)U^(235))=235.043933 "amu" m(._(0)n^(1))=1.008665 amu m(._(56)Ba^(141))=140.917700 "amu" m(._(36)Kr^(92))=91.895400 amu

Assuming that in a star, three alpha particle join in a single fusion reaction to form ._(6)C^(12) nucleus. Calculate the energy released in this reaction. Given mass of ._(2)He^(4) is 4.002604 a.m.u. and that of ._(6)C^(12) is 12 a.m.u. Take 1a.m.u. =931MeV.

Assuming that four hydrogen atom combine to form a helium atom and two positrons, each of mass 0.00549u, calculate the energy released. Given m(._(1)H^(1))=1.007825u, m(._(2)He^(4))=4.002604u .

Calculate the binding energy per nucleon of ._20Ca^(40) nucleus. Given m ._20Ca^(40)=39.962589u , M_(p) =1.007825u and M_(n)=1.008665 u and Take 1u=931MeV .

Calculate the energy of the following nuclear reaction: ._(1)H^(2)+._(1)H^(3) to ._(2)He^(4) + ._(0)n^(1)+Q Given: m(._(1)H^(2))=2.014102u, m(._(1)H^(3))=3.016049u, m(._(2)He^(4))=4.002603u, m(._(0)n^(1))=1.008665u

A nucleus X, initially at rest, undergoes alpha-decay according to the equation. _92^AXrarr_Z^228Y+alpha (a) Find the values of A and Z in the above process. (b) The alpha particle produced in the above process is found in move in a circular track of during the process and the binding energy of the parent nucleus X. Given that m(Y)=228.03u , m( _0^1n)=1.009u m( _2^4He)=4.003u , m( _1^1H)=1.008u