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Physics 1 Problem Solutions 296

# Physics 1 Problem Solutions 296 - not even suspected for...

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292 Chapt. 39 Photons and Matter Waves expected the alpha particles to pass right through the foils with only small deviations. Most did, but some scattered off a very large angles. Using a classical particle picture of the alpha particles and the entities they were scattering off of he came to the conclusion that atoms contained most of their mass and positive charge inside a region with a size scale of 10 15 m = 1 fm: this 10 5 times smaller than the atomic size. (Note fm stands officially for femtometer, but physicists call this unit a fermi.) Rutherford concluded that there must be a dense little core to an atom: the nucleus. a) Why did the alpha particles scatter off the nucleus, but not off the electrons? HINTS: Think dense core and diffuse cloud. What is the force causing the scattering? b) If the alpha particles have kinetic energy 7 . 5 Mev, what is their de Broglie wavelength? c) The closest approach of the alpha particles to the nucleus was of order 30 fm. Would the wave nature of the alpha particles have had any effect? Note the wave-particle duality was
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Unformatted text preview: not even suspected for the massive particles in 1911. 039 qfull 00600 2 3 0 moderate math: complex numbers Extra keywords: (HRW-977:66E) 3. A complex number z is de²ned by z = x + iy , where x and y are real numbers and i = √ − 1 is imaginary. The real part of z is x = Re[ z ] and the imaginary part is y = Im[ z ] The complex conjugate of z is de²ned by z ∗ = x − iy . a) What is zz ∗ ? b) Prove that ( z 1 z 2 ) ∗ = z ∗ 1 z ∗ 2 . c) Prove that p 1 z P ∗ = 1 z ∗ . d) Prove that p z 1 z 2 P ∗ = z ∗ 1 z ∗ 2 . 039 qfull 00700 2 5 0 moderate thinking: Schr¨odinger eqn. 4. Do the following. a) Verify that the full time-dependent wave function Ψ( x,t ) = ψ ( x ) e − iEt/h − reduces the full time-dependent Schr¨odinger equation − h − 2 2 m ∂ 2 Ψ ∂x 2 + V ( x )Ψ = ih − ∂ Ψ ∂t to the time-independent Schr¨odinger equation − h − 2 2 m ∂ 2 ψ ∂x 2 + V ( x ) ψ = Eψ...
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