p41_004 - m l = l . Thus [ L z ] max = l ¯ h = 3(1 . 055...

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4. (a) We use Eq. 41-2: L = p l ( l +1)¯ h = p 3(3 + 1) (1 . 055 × 10 34 J · s) = 3 . 653 × 10 34 J · s . (b) We use Eq. 41-7: L z = m l ¯ h . For the maximum value of L z set
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Unformatted text preview: m l = l . Thus [ L z ] max = l ¯ h = 3(1 . 055 × 10 − 34 J · s) = 3 . 165 × 10 − 34 J · s ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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