quiz 3 solution

# quiz 3 solution - PHYS126 Quiz 3 Student Name Student ID...

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Unformatted text preview: PHYS126 Quiz 3 17-3-2011 Student Name: Student ID: Time: 20min Note: You must answer each question by putting A, B, C, D or E in the boxes provided below. Marks will NOT be given by other ways of answering the question such as circling, crossing out the choices or write the answer in somewhere else. Physical constants provided: Planck constant: 34 6.63 10 Js h- = × Speed of light: 8-1 3 10 ms c = × Electronic charge: 19 1.6 10 C e- = ×------------------------------------------------------------------------------------------------------- 1. The energy levels of the hydrogen atom are given in terms of the principal quantum number n and a positive constant A by the expression (A) 1 2 A n + (B) 2 An (C) 2 / A n (D) 2 An- (E) 2 / A n- Answer: Answer: The energy levels of the hydrogen atom is 2 E E n = - , where 4 2 2 13.6eV 8 me E h ε = = is a positive number. Therefore the answer is (E).------------------------------------------------------------------------------------------------------- 2. In the spectrum of hydrogen, what is the ratio of the longest wavelength in the Lyman series 1 n = , to the longest wavelength in the Balmer series 2 n = ? (A) 5/27 (B) 2/3 (C) 1/4 (D) 3/2 (E) 27/5 Answer: Answer: Lyman series is the series of transitions as an electron goes from any upper energy state of 1 n to 1 n = , i.e. 2 2 2 1 1 1 L E E E E n n ∆ = -- - =- where L L E h ν ∆ = is also the energy of the emitted photon. E A 2 2 1 1 1 1 1 L L L E hc h E n hc n ν λ λ = =- ⇒ =- Since n must be an integer, the longest wavelength (or lowest frequency) in Lyman series is 2 1 1 1 2 L E hc λ =- . Similarly, Balmer series is the series of transitions as an electron goes from any upper energy state of 2 n to 2 n = , i.e....
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quiz 3 solution - PHYS126 Quiz 3 Student Name Student ID...

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