Question

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 1

Correct Answer - `[(a) ((q_(alpha)E_(0))/m_(alpha)t)i+v_(0) cos theta hat(j)-v_(0) sin theta hat(k)` where `theta=omegat` and `omega=(q_(alpha)B)/m_(alpha)`, (b) `10^(7) m//s` (c) `8.00 MeV]`