CH 39 #37 - 39.37 a)  E = hc = 12eV b) Find E for an...

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Unformatted text preview: 39.37: a)  E = hc λ = 12eV b) Find E for an electron with λ = 0.10 × 10 −6 m. λ = h p so p = h λ = 6.626 × 10 −27 kg ⋅ m s E = p 2 (2m) = 1.5 × 10 −4 eV E = q∆V so ∆V = 1.5 × 10 −4 V v = p m = (6.626 × 10 −27 kg ⋅ m s) (9.109 × 10 −31 kg) = 7.3 × 103 m s c) Same λ so same p. E = p 2 (2m) but now m = 1.673 × 10 −27 kg so E = 8.2 × 10 −8 eVand ∆V = 8.2 × 10 −8 V. v = p m = (6.626 × 10−27 kg ⋅ m s) (1.673 × 10−27 kg) = 4.0 m s ...
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This homework help was uploaded on 04/02/2008 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.

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