A neutron of kinetic energy 65 eV collides inelastically with a singly ionised helium atom at rest. It is scattered at an angle `90^@` with respect to original direction. If the energy of scattered neutron is 6.36 eV, find the frequency of emitted radiation from the helium atom after suffering collision.

Answer

(a) `6.36eV, 0.312eV` (of neutron), `17.84eV, 16.328eV` (of atom)
(b) `1.82xx10^(15)Hz, 11.67xx10^(15)Hz, 9.84xx10^(15)`

Explanation

Let `K_(1)` and `K_(2)` be the kinectic energies of neutron and helium atom after collision and `DeltaE` be the excitation energy



From conservation of linear momentum along x-direction.
`rArr sqrt(2Km)=sqrt(2(4m)K_(2))cos theta(P=sqrt(2Km))`
Similarly.,applying conservation of linear momentum in y-direction, we have
`sqrt(2m_(1)m) = sqrt(2(4m)K_(2))sin theta`
Squaring and adding equation (1) and (2), we get
`K+K=4K_(2)`
or `4K_(2)-K_(1)=K=65 eV`
Now, during collision, electron cana be excited to any higher energy state. Applying conservation of energy we get `K=K_(1)+K_(2)+Delta`

or `65=K_(1)+K_(2)+DeltaE`
`DeltaE` can have the following values,
`DeltaE_(1)={-13.6-(-54.4)}eV=40.8 eV`
Substituting in `(5)`, we get
`K_(1)+K_(2)=24.2eV`
Solving (4) and (6), we get
`K_(1)=6.36 eV` and `K_(2)=17.84 eV`
Similarly, when we put `DeltaE=DeltaE_(2)`
`{-6.04-(54.4)}eV`
in equation `(5)`, we get
`K_(1)+K_(2)=14 eV`
Now, solving `(4)` and `(8)`, we get
`K_(1)= -1.8 eV` and `K_(2)=15.8 eV`
But since the Kinetic energies cannot have the values, the electron will not jump to third excited state of `n=4`.
Therefore, the allowed values of `K_(1)` ( KE of neutron) are `6.36 eV` and `0.312 eV` and of `K_(2)` (KE of the atom) are `17.84 eV` and `16.238 eV` and the electron can jump upto second exicted state only `(n=3)`
(b) Possible emission lines are only three as shown in figure. The correspinding frequencies are
`V_(1)=((E_(3)-E_(2)))/(h)`
`=({-6.04-(-13.6)}xx1.6xx10^(-19))/(6.63xx10^(-34))Hz`
`=1.82xx10^(15)Hz`
`V_(2)=(E_(2)-E_(1))/(h)`
`=({-6.04-(54.4)}xx1.6xx10^(1-9))/(6.63xx10^(-34))Hz`
`11.67xx10^(15)Hz`
and `V_(3)=(E_(2)-E_(1))/(h)`
`=({-13.6-(54.4)}xx1.6xx10^(-19))/(6.63xx10^(-34))Hz`
`9.84xx10^(15)Hz`
Hence, the frequencies of emitted radiations are `1.82xx10^(5)Hz, 11.67xx10^(15)Hz` and `9.84xx10^(15)Hz`.