A neutron of kinetic `6.5 eV` collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle `90^(@)` with respect to its original direction.
Find the maximum allowed value of energy of the He atom?

Options

A. `13.68 eV`
B. `19.88 eV`
C. `15.26 eV`
D. `17.84 eV`

Answer

D

Explanation

Let the final speed of nature and singly ionization helium atom be `v_(1)` and `v_(2)`, respectivelly. From conservation of momentum , we have
Along x-axis: `mu = 4 mv_(2) cos theta` (i)
Along y-axis: `mu_(1) = 4 mv_(2) sin theta` (ii)
On squaring and adding equations (i) and (ii) , we get `u^(2) + v_(1)^(2) = 16v_(2)^(2)`
`(1)/(2) mu^(2) + (1)/(2) mv_(1)^(2) = (1)/(2) 16mv_(2)^(2)` (iii)
The initial kinetic energy of neutron `= 65 eV`.
Let kinetic energy of neutron and helium atom be
`K_(n) = (1)/(2) mv_(1)^(2),K_(He) = (1)/(2) (4m)v_(2)^(2)`
So, Eq. (iii) reduces to `65 + K_(n) = 4K_(He)`
`4K_(He) - K_(n) = 65` (iv)
a. Energy required to excite an electron from ground state of singly ionized helium atom `(He^(+))` to `n^(th)` energy level is `u^(2) + v_(1)^(2) = 16 v_(2)^(2)`
If the neutron has cufficient energy to excite the helium atom, then from conservation of energy , the energy of neutron must be equal to the sum of kinetic energy of neutron, helium atom, and excitation energy. So, we have
`65 = K_(n) + K_(He) + 54.4 (1 - (1)/(n^(2))`
`K_(He) + K_(n) = 10.6 + (54.4)/(n^(2))` (v)
On solving Eqs, (iv) and (v), we get
`K_(He) = [15.52 + (10.88)/n^(2)] eV` (vi)
`K_(n) = [(43.52)/(n^(2)) - 4.52] eV` (vii)
The kinetic energy is always positive, so from equation (vii), we have
`(43.52)/(n^(2)) gt 4.52. n lt 3.1`
So, the only possible values of `n` are `2` and `3` Possible values of `n`
`2,3`
Allowed value of neutron energy
`K_(N)`
`6.36 eV, 0.32 eV`
Allowed value of He atom energy
`K_(He)`
`17.84 eV, 16.33 eV`
b. When the atom de-excites,
`v = ((13.6) 2^(2) xx 1.6 xx 10^(-19))/(6.63 xx 10^(-34)) [(1)/(n_(l)^(2)) - (1)/(n_(u)^(2))]`
`= 13.3 [(1)/(n_(l)^(2)) - (1)/(n_(u)^(2))] xx 10^(15) Hz`
The electron excited to `n = 3` can make three transition:
`n = 3` to `n = 1, v_(1) = 11.67 xx 10^(15) Hz`
`n = 3` to `n = 2, v_(2) = 9.84 xx 10^(15) Hz`
`n = 2` to `n = 1, v_(3) = 1.83 xx 10^(15) Hz`