problem40_16 - 10 38 1 s J 10 63(5.375(6 m/s 10 00 3 m 10...

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40.16: Since = E U 6 0 we can use the result = E E 625 . 0 1 from Section 40.3, so = - E E U 375 . 5 1 0 and the maximum wavelength of the photon would be h c mL mL h hc E U hc ) 375 . 5 ( 8 ) 8 )( 375 . 5 ( λ 2 2 2 1 0 = = - = m.
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Unformatted text preview: 10 38 . 1 s) J 10 63 (5.375)(6. m/s) 10 00 . 3 ( m) 10 kg)(1.50 10 11 . 9 ( 8 6 34 8 2 9 31----× = ⋅ × × × × =...
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