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]`.

Options

A. `1.2xx10^(7) m//s`
B. `9xx10^(6) m//s`
C. `1.5xx10^(7) m//s`
D. `8xx10^(5) m//s`

Answer

C

Explanation

`R=(mv sin theta)/(qB)rArrv sin theta=1.2xx10^(7)m//s` (Charge of a-particle `=3.2'10^(-19)C`)
`P=(2 pi m)/(qB)v cos thetarArrv cos theta=(pqB)/(2 pi m)=9xx10^(6)m//s`
`:. V_(alpha)=sqrt((v sin theta)^(2)+(v cos theta)^(2))=1.5xx10^(7)m//s`