A `He^(+)` ion is at rest and is in ground state. A necutron with intial kinetic energy`K` collides heads on with the `He^(+)` ion. Find minimum value of `K` so that there can be an inelastic collision between these two particals.

Options

A. `K_(min) = 51 e V`
B. `K_(min) = 48.6 e V`
C. `K_(min) = 512 e V`
D. `K_(min) = 13.6 e V`

Answer

A

Explanation

Here the loss during the collision can only be used to excite the atoms or electrons
So , according to quantum mechaincs possible losses can be
`{0, 40.8 e V, 48.36 e V}` (i)
from `E_(n) = - 13.6 e V (Z^(2))/(n^(2))`
Now according to Newtonion mechanics. Minimum loss `= 0`. Maximum loss will be for perfectly inelclastic collision.
Let `v_(0)` be the inital speed of neutron and `v_(f)` be the final common speed
So, by momentum conservation :
`m v_(0) = m v_(f) + 4m v_(f) implies v_(f) = (v_(0))/(5)`
where `m = "mass of meutron and mass of" He^(+)` ion =` 4 m`. Final kinetic energy of system:
`KE = (1)/(2) m v_(f)^(2) + (1)/(2) 4 m v_(f)^(2)`
`= (1)/(2) (5 m) = v_(0)^(2)/(25) = (1)/(5) ((1)/(2) m v_(0)^(2)) = (K)/(5)`
maximum loss `= K - (K)/(5) = (4K)/(5)`
So , loss will be `[0 (4K)/(5)]`
For inelastic collition, there should be at least one common value other than zero in Eqs. (i) and (ii)
`:. (4K)/(5) gt 40.8 e V implies K gt 51 e V`
Hence maximum value of `K` for an inelastic collision , `K_(min) = 51 e V`.