Question

A monoenergetic beam a alpha particle , each of them having kinetic E is incident on a sample of singly ionised Helium gas in ground state . The Helium gas in ground state. The Helium sample may start emitting radiation Since the Kinetic energy of incident alpha particle in sufficiently high . A part of this radiation (if any) is allowed to pass through atomic Hydrogen sample in ground state. A detector placed near the hydrogen records both radiation and electrons. Assume all `He^(+)` ions to be initially at rest.
If maximum Kinetic energy of electrons interpected by the detector is `27.2eV`, the minimum kinetic of alpha paticle energy of alpha particle ,must be-
A. `81.6eV`
B. `96.8eV`
C. `102eV`
D. 40.8eV`

Answer 1

Correct Answer - A
Maximum energy of photon incident in H-sample =`13.6eV+27.2=40.8eV`
This corresponding to the transition `2` to `1` in `He^(+)` sample for this electrons in `He^(+)` sample must be excited to a maximum of 1 st excited state. Since mass of alpha-particle and `He^(+)` ion is almost same , the alpha -particle transfer `1//2` of their kinetic energy to excitation energy `(40.8eV)` for `He^(+)` ion.
` KE_(alpha)=2xx1^(st)` excitation energy of `He^(+)= 2xx40.8=81.6eV`