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Early Models of the Atom Contents: J. J. Thomson The electron Plum pudding model of the atomPowerPoint Presentation

Early Models of the Atom Contents: J. J. Thomson The electron Plum pudding model of the atom

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- Early Models of the Atom
- Contents:
- J. J. Thomson
- The electron
- Plum pudding model of the atom

- Rutherford
- Gold foil experiment
- Nucleus model
- Solving closest approach problems
- Whiteboard

- J. J. Thomson

J.J. Thomson 1856 - 1940

“Plum Pudding” model

Discovers the electron (charge/mass ratio Straight/curved)

Electrons are negatively charged

Electrons are surrounded in atom with positive “stuff”

TOC

Ernest Rutherford(1871-1937)

Sends alpha particles into gold foil

Measures the angle of scatter

Should go right through plum pudding…

TOC

Ernest Rutherford(1837-1937)

Rutherford’s atom:

(It has a nucleus)

- But is also has problems:
- Why doesn’t the electron radiate energy?
- How does this explain the spectral lines they had been observing?
- I won’t Bohr you with the solution to this right now…

TOC

Most Alphas are not deflected much

More deflection closer to nuclei

Ek = 1/2mv2

V = W/Q, W = VQ

V = kQ/r

VQp = 1/2mv2

kinetic = potential

1/2mv2 = (kQN/r)Qp

r

QN

Qp

Example 1: What is the closest approach of a proton going 2.6 x 106 m/s if it approaches a carbon nucleus head on?

TOC

What is the closest approach in nm of a proton going 15,000 m/s to a hydrogen nucleus.

Mp = 1.673 x 10-27 kg

A hydrogen nucleus has 1 proton in it

kinetic energy = electric potential energy

Ek = 1/2mv2 = 1/2(1.673 x 10-27 kg)(15000 m/s)2 = 1.88213 x 10-19 J

Ek = electric potential energy = VQ = Ve

Ek = (kQ/r)e = (ke/r)e = ke2/r, r = ke2/Ek

= (9.0 x 109 Nm2/C2)(1.602 x 10-19 C)2/(1.88213 x 10-19 J)

= 1.22721 x 10-9m = 1.2 nm

W

1.2 nm

An Alpha particle’s closest approach brings it to within 61 nm of a Gold nucleus. What is its velocity in m/s? (3)

Mp = 1.673 x 10-27 kg, Mn = 1.675 x 10-27 kg, = 2n+2p

M = 6.696 x 10-27 kg

A gold nucleus has 79 protons in it, and an alpha 2

kinetic energy = electric potential energy

electric potential energy = VQ = V2e (alpha)

V = kQ/r = (9.0 x 109 Nm2/C2)(79*1.602 x 10-19 C)/(61 x 10-9 m)

= 1.867 V

elec. pot. = V2e = (1.867 V)(2)(1.602 x 10-19 C) = 5.983 x 10-19 J

5.983 x 10-19 J = 1/2mv2 = 1/2(6.696 x 10-27 kg)v2

v = 13367 m/s = 13,000 m/s

W

13,000 m/s

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