Fundamentals Of Physics - Volume 1 Only
11th Edition
ISBN: 9781119306856
Author: Halliday
Publisher: WILEY
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Chapter 39, Problem 19P
To determine
To find:
The energy of the first excited state of the electron
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The light observed that is emitted by a hydrogen atom is explained by a simple model of its structure with one proton in its nucleus and an electron bound to it, but only with internal energies of the atom satisfying
EH=−RH/n2EH=−RH/n2
where RHRH is the Rydberg constant and nn is an integer such as 1, 2, 3 ... and so on. When a hydrogen atom in an excited state emits light, the photon carries away energy and the atom goes into a lower energy state.
Be careful about units. The Rydberg constant in eV is
13.605693009 eV
That would be multiplied by the charge on the electron 1.602× 10-19 C to give
2.18× 10-18 J
A photon with this energy would have a frequency f such that E=hf. Its wavelength would be λ = c/f = hc/E. Sometimes it is handy to measure the Rydberg constant in units of 1/length for this reason. You may see it given as 109737 cm-1 if you search the web, so be aware that's not joules.
The following questions are intended to help you understand the connection between…
The x-ray spectrum is for 35.0 keV electrons striking a molybdenum (Z= 42) target. If you substitute a silver (Z = 47) target for the molybdenum target, will (a) lmin, (b) the wavelength for the Ka line, and (c) the wavelength for the Kb line increase, decrease, or remain unchanged?
The longest wavelength line of the Balmer Series for hydrogen occurs at 656.3 nm corresponding to the transition from n2 = 3 to n1 = 2. Determine the value for the Rydberg constant for hydrogen using these values.
Chapter 39 Solutions
Fundamentals Of Physics - Volume 1 Only
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