Suppose a nucleus initally at rest undergoes `alpha` decay according to equation
`._(92)^(225)X rarr Y +alpha`
At `t=0`, the emitted `alpha`-particles enter a region of space where a uniform magnetic field `vec(B)=B_(0) hat i` and elecrtic field `vec(E)=E_(0) hati` exist. The `alpha`-particles enters in the region with velocity `vec(V)=v_(0)hat j` from `x=0`. At time`t=sqrt3xx10^(6) (m_(0))/(q_(0)E_0)s`, the particle was observed to have speed twice the initial velocity `v_(0)`. Then, find (a) the velocity `v_(0)` of the `alpha`-particles,
(b) the initial velocity `v_(0)` of the `alpha`-particle, (c ) the binding energy per nucleon of the `alpha`-particle.
`["Given that" m(Y)=221.03 u,m(alpha)=4.003 u,m(n)=1.09u,m(P)=1.008u]`.

Answer

(a) `v_(0)=10^(7)m//s`
(b) `omega=qB//m`
(c )`8.11 MeV//"nucleon"`

Explanation

(a) `(2V_(0))^(2)=((qeE)/(m)xxsqrt(3)xx10^(7)(m)/(qE))^(2)+v_(0)^(2)rArrv_(0)=10^(7)m//s`
(b) The path will be helical. `vec(v)=(qE)/(m)t hat(i)+v_(0)cos theta hat(j)-v_(0)sin theta hat(k)`
`:. vec(v)(t)=(qE_(0))/(m)t hat (i)+10^(7)cos omegahat (j)-10^(7) sin omega t hat (k)` where `omega=qB//m`
(c ) `k_(alpha)=(M_y)/(M_(y)+M_(alpha))Q rArrQ(225)/(221)kalpha=(225)/(221)xx(1)/(2)m_(alpha)v_(0)^(2)`
`Q=Delta mc^(2)rArr(225)/(221)xx(1)/(2)m_(alpha)v_(0)^(2)=Deltamc^(2)`
`Deltam=(225)/(2xx221)m_(alpha)(V_(0)^(2))/(c^(2))=(225)/(2xx221)xx4xx(10^(14))/(9xx10^(16))=(50)/(221xx10^(2))=(1)/(442)a.m.u`
Now, `m_(x)=m_(y)+m_(alpha)+Deltam=225.033+(1)/(442)`
`BE=[192m_(p)+133m_(n)-m_(x)]c^(2)=1.96xx931MeV`
`:. BE` per nucleon `=(1.96xx931)/(225) MeV//"nucleon"=8.11MeV//"nucleon"`