An Introduction to Physical Science
14th Edition
ISBN: 9781305079120
Author: James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Publisher: Brooks Cole
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Chapter 9, Problem 9E
To determine
Energy in eV of the photon emitted when an electron jumps down from the
n = 4
orbit to
n = 2
orbit of a hydrogen atom.
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A hypothetical atom has only two atomic energy levels, separated by 3.2 eV. Suppose that at a certain altitude in the atmosphere of a star there are 6.1 * 1013/cm3 of these atoms in the higher-energy state and 2.5 * 1015/cm3 in the lower-energy state. What is the temperature of the star’s atmosphere at that altitude?
A photoelectron is emitted from K shell (n = 1) of a carbon atom, and an election in L shell (n = 2) moves down to the vacancy in K shell. What is the wavelength, in the unit of nm, of the photon emitted during this transition?
Use for the energy difference between two states in an atom.
E0 = 13.6 eV and atomic number of carbon is Z = 12.
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It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.
Chapter 9 Solutions
An Introduction to Physical Science
Ch. 9.1 - Prob. 1PQCh. 9.1 - Prob. 2PQCh. 9.2 - Prob. 1PQCh. 9.2 - Prob. 2PQCh. 9.2 - Prob. 9.1CECh. 9.3 - Prob. 1PQCh. 9.3 - When does a hydrogen atom emit or absorb radiant...Ch. 9.3 - Prob. 9.2CECh. 9.3 - Prob. 9.3CECh. 9.3 - Prob. 9.4CE
Ch. 9.4 - Prob. 1PQCh. 9.4 - Prob. 2PQCh. 9.5 - Prob. 1PQCh. 9.5 - Prob. 2PQCh. 9.6 - Prob. 1PQCh. 9.6 - Prob. 2PQCh. 9.6 - Prob. 9.5CECh. 9.7 - Prob. 1PQCh. 9.7 - Prob. 2PQCh. 9 - Prob. AMCh. 9 - Prob. BMCh. 9 - Prob. CMCh. 9 - Prob. DMCh. 9 - Prob. EMCh. 9 - Prob. FMCh. 9 - Prob. GMCh. 9 - Prob. HMCh. 9 - Prob. IMCh. 9 - Prob. JMCh. 9 - Prob. KMCh. 9 - Prob. LMCh. 9 - Prob. MMCh. 9 - Prob. NMCh. 9 - Prob. OMCh. 9 - Prob. PMCh. 9 - Prob. QMCh. 9 - Prob. 1MCCh. 9 - Prob. 2MCCh. 9 - Prob. 3MCCh. 9 - Prob. 4MCCh. 9 - Prob. 5MCCh. 9 - Prob. 6MCCh. 9 - Prob. 7MCCh. 9 - Prob. 8MCCh. 9 - Prob. 9MCCh. 9 - Prob. 10MCCh. 9 - Prob. 11MCCh. 9 - Prob. 12MCCh. 9 - Prob. 13MCCh. 9 - Prob. 14MCCh. 9 - Prob. 1FIBCh. 9 - Prob. 2FIBCh. 9 - Prob. 3FIBCh. 9 - Prob. 4FIBCh. 9 - Prob. 5FIBCh. 9 - Prob. 6FIBCh. 9 - Prob. 7FIBCh. 9 - Prob. 8FIBCh. 9 - Prob. 9FIBCh. 9 - Prob. 10FIBCh. 9 - Prob. 11FIBCh. 9 - Prob. 12FIBCh. 9 - Prob. 1SACh. 9 - Prob. 2SACh. 9 - Prob. 3SACh. 9 - Prob. 4SACh. 9 - Prob. 5SACh. 9 - Prob. 6SACh. 9 - Prob. 7SACh. 9 - Prob. 8SACh. 9 - Prob. 9SACh. 9 - Prob. 10SACh. 9 - Prob. 11SACh. 9 - Prob. 12SACh. 9 - Prob. 13SACh. 9 - Prob. 14SACh. 9 - Prob. 15SACh. 9 - Prob. 16SACh. 9 - Prob. 17SACh. 9 - Prob. 18SACh. 9 - Prob. 19SACh. 9 - Prob. 20SACh. 9 - Prob. 21SACh. 9 - Prob. 22SACh. 9 - Prob. 23SACh. 9 - Prob. 24SACh. 9 - Prob. 25SACh. 9 - Prob. 26SACh. 9 - Prob. 27SACh. 9 - Prob. 28SACh. 9 - Prob. 29SACh. 9 - Prob. 30SACh. 9 - Prob. 31SACh. 9 - Prob. 32SACh. 9 - Prob. 33SACh. 9 - Prob. 34SACh. 9 - Visualize the connection for the descriptions of...Ch. 9 - Prob. 1AYKCh. 9 - Prob. 2AYKCh. 9 - Prob. 3AYKCh. 9 - Prob. 4AYKCh. 9 - Prob. 5AYKCh. 9 - Prob. 1ECh. 9 - Prob. 2ECh. 9 - Prob. 3ECh. 9 - Prob. 4ECh. 9 - Prob. 5ECh. 9 - Prob. 6ECh. 9 - Prob. 7ECh. 9 - Prob. 8ECh. 9 - Prob. 9ECh. 9 - Prob. 10ECh. 9 - Prob. 11ECh. 9 - Prob. 12E
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- A neutral sodium atom has an ionization potential of 5.1 eV. What is the speed of a free electron that has just barely enough kinetic energy to collisionally ionize a sodium atom in its ground state? What is the speed of a free proton with just enough kinetic energy to collisionally ionize this atom?arrow_forwardThe 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…arrow_forwardA triply ionised beryllium atom (Be+++, Z = 4) has only one electron in orbit about the nucleus. If the electron decays from the n 7 level to the first excited state (n = 2), calculate the wavelength of the photon emitted. Please give your answer in units of nm, rounded to one decimal place. Answer:arrow_forward
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