1182_Physics ProblemsTechnical Physics

# 1182_Physics ProblemsTechnical Physics - Chapter...

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Chapter 42 523 * P42.14 (a) 11 1 2 22 λ =− F H G I K J ZR nn H fi . The shortest wavelength, s , corresponds to n i =∞ , and the longest wavelength, A , to if =+ 1 . 1 2 2 s H f n = (1) 1 1 1 1 2 2 2 2 A + L N M M M O Q P P P + F H G I K J L N M M O Q P P n n n n n H f f H f f f di (2) Divide (1) and (2): s f f n n A + F H G I K J 1 1 2 + = n n f f s 1 22 8 0800 A . . nm 63.3 nm ∴= n f 4 From (1): Z n R f sH == ×× = −− 2 2 97 1 4 22 8 10 1 097 10 800 .. . m m ej e j . Hence the ion is O 7 + . (b) + L N M M O Q P P R S | T | U V | W | 70208 10 1 4 1 4 81 1 . m a f k , k = 123 ,,, Setting k = 234 , , gives = 41 0 . nm, 33.8 nm, 30.4 nm . P42.15 (a) The speed of the moon in its orbit is v r T × × 2 2 3 84 10 236 10 102 10 8 6 3 π . . . m s ms . So, Lm v r × × × = × 7 36 10 1 02 10 3 84 10 2 89 10 22 3 8 34 . . kg m s m kg m s 2 e j e j . (b) We have
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