hw8_solutions

# hw8_solutions - PHY 251 Fall 2009 homework problem set 8...

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Unformatted text preview: PHY 251 Fall 2009: homework problem set 8, due in the PHY 251 drop box in room A-129 by noon on Friday, Nov. 13. 1. Serway 8.7 Answer: This problem is separable, in that one can solve the Schr¨odinger equation for each dimension separately. Let’s assume ψ ( x, y, z ) = A sin( n 1 π x L 1 ) sin( n 2 π y L 2 ) sin( n 3 π z L 3 ) and integrate to normalize: 1 = A 2 [ integraldisplay L 1 x =0 sin 2 ( n 1 π x L 1 ) dx ] [ integraldisplay L 2 y =0 sin 2 ( n 2 π y L 2 ) dy ] [ integraldisplay L 3 z =0 sin 2 ( n 3 π z L 3 ) dz ] Now all three integrals have identical form. Let’s solve for x , using a = n 1 π/L 1 : integraldisplay L 1 x =0 sin 2 ( ax ) dx = [ x 2 − sin(2 ax ) 4 a ] vextendsingle vextendsingle vextendsingle vextendsingle x = L 1 x =0 = [ L 1 2 − sin(2 n 1 πL 1 /L 1 ) 4 n 1 π/L 1 ] − [ 2 − sin(2 n 1 π /L 1 ) 4 n 1 π/L 1 ] = [ L 1 2 − 0] − [0 − 0] = L 1 2 because sin(2 n 1 π ) = 0 and sin(0) = 0 . We thus have 1 = A 2 L 1 2 L 2 2 L 3 2 or A = radicalBig 8 / ( L 1 L 2 L 3 ) . 2. Write out the full wavefunction ψ for the 2 p state of hydrogen with L z = − ¯ h . What’s the Bohr model energy for this state? Answer: The 2 p state with L z = − ¯ h has n = 2 , ℓ = 1 , and m ℓ = − 1 , so the wavefunction is ψ = R n =2 ,ℓ =1 ( r ) Y m ℓ = − 1 ℓ =1 ( θ, φ ) = parenleftbigg Z 2 a parenrightbigg 3 / 2 Zr √ 3 a e − Zr/ 2 a − 1 2 radicalBigg 3 2 π sin( θ ) e + iφ where we have used Table 8.4 on p. 280 for R nℓ ( r ) and Table 8.3 on p. 269 for Y m ℓ ℓ ( θ, φ ) . 3. Calculate numerical values for the net angular momentum, and for all possible ˆ z axis angular momenta, for the 2 p and 4 d states of hydrogen. Answer: The 2 p state has n = 2 and ℓ = 1 , with m ℓ = − 1 , , +1 . Numerical values are | L | = ¯ h radicalBig ℓ ( ℓ + 1) = (1 . 055 × 10 − 34 J · s ) radicalBig 1(1 + 1) = 1 . 492 × 10 − 34 J · s and L z = {− 1 , , 1 } · 1 . 055 × 10 − 34 J · s. The 4 d state has n = 4 and ℓ = 2 with m ℓ = − 2 , − 1 , , 1 , 2 , so | L | = ¯ h radicalBig ℓ ( ℓ + 1) = (1 . 055 × 10 − 34 J · s ) radicalBig 2(2 + 1) = 2 . 584 × 10 − 34 J · s and L z = {− 2 , − 1 , , 1 , 2 } · (1 . 055 × 10 − 34 J · s ) ....
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hw8_solutions - PHY 251 Fall 2009 homework problem set 8...

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