Atoms Flashcards
graph of scattering angle with no. of α particles scattered
kind of like inversely proportional
Significance of α particle experiment
Most of them passed undeflected
Most space of atom is empty
Few deflected through 90 & 180
All +ve charge & the mass of atom is concentrated in a very small region called nucleus
Size of nucleus is 1/10^15m & that of atom is 1/10^10m
Since gold nucleus is much heavier than α particle, gold nucleus remains at rest
Impact parameter
Perpendicular distance of initial velocity vector of α particle from center of nucleus
For large impact parameter, α particle is
almost undeflected
For head on collision, impact parameter is
0
Rutherford’s postulates
Entire +ve charge & mass of atom is concentrated at very small volume at the center called nucleus
Amount of +ve charge is equal to -ve charge
Atom is neutral
E’s are continuously revolving around the nucleus in circular orbits
Force of attraction provides necessary centripetal force
Limitations of Rutherford’s atomic model
- an accelerating particle emits radiation as EM waves. So energy of the electron show gradually decrease leading to decrease in radius of e’ orbit
- Electron follows a spiral path & finally fall into the nucleus so atom isn’t stable
- As e’ spiral inwards, their angular velocity & their freq change continuously they would emit continuous spectrum instead of actually observed line spectrum
Bohr’s atomic model
- In a H atom, an e’ revolves in stable orbits called stationary orbits
- These orbits are those for which angular momentum is integral multiple of h/2pi
As long as e’ is revolving in stationary orbit, it doesn’t radiate energy. When e’ make a transition from higher energy to lower, a photon is emitted, having energy equal to energy diff b/n initial & final states
radius of orbit r =
n²h²ε/Ze²pi m
Are stationary orbits equally spaced?
No
Speed _______ as we move away from the nucleus
decreases
Potential energy U =
-kZe²/r
Kinetic energy K =
kZe²/2r
E =
-mZ²e⁴/8ε²n²h²
-ve sign in E indicates that?
e’ is bound to nucleus so energy is needed to be done to separate e’ from nucleus