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Find the maximum fractional energy loss for a 0.511-MeV

Physics for Scientists and Engineers with Modern Physics | 9th Edition | ISBN: 9781133954057 | Authors: Raymond A. Serway John W. Jewett ISBN: 9781133954057 230

Solution for problem 36 Chapter 40

Physics for Scientists and Engineers with Modern Physics | 9th Edition

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Physics for Scientists and Engineers with Modern Physics | 9th Edition | ISBN: 9781133954057 | Authors: Raymond A. Serway John W. Jewett

Physics for Scientists and Engineers with Modern Physics | 9th Edition

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Problem 36

Find the maximum fractional energy loss for a 0.511-MeV gamma ray that is Compton scattered from (a) a free electron and (b) a free proton.

Step-by-Step Solution:
Step 1 of 3

CHM 170 Chapter 18 part 2 The Half Reaction Method in an acidic medium (As demonstrated by this equation) SO +2Cr O 2 72-------------- SO42-+ Cr 3+ First assign oxidation values: +4 -2 +6 -2 +6 -2 +3 S O 2 +Cr 2 O7------- S O 4 + Cr Next split the reaction into two half reactions: +4 -2 +6 -2 Oxidation: S O 2 --------- S O 4 +6 -2 +3 Reduction: Cr 2 O 7 -------- Cr Now we have to balance the atoms do this by adding an H O for every Oxygen atom 2 and 2 hydrogens on the opposite side to balance out the equation like so: Oxidation: SO + 22 O ----2------------- SO 4-2+ 4H + -2 + +3 Reduction: Cr O 2 7 + 14H -------------- 2Cr + 7H O 2 Add electrons to make the charges even: Oxidation: SO + 22 O ----2--------------- SO 4-2+ 4H + 2e - Reduction: Cr O 2 7-2 + 14H + 6e ---------------- 2Cr +3 + 7H O 2 Next we use multiplication to make the electrons equal: -2 + - 3[SO +22H O --2----------- SO 4 + 4H + 2e ] 3SO + 6H O --------------- 3SO -2+ 12H + 6e - 2 2 4 Finally we combine the equations: -2 + - 3SO +26H O --2-- 3SO 4 + 12H + 6e -2 + - +3 + Cr O2 7 + 14H + 6e ------ 2Cr + 7H O 2 3SO + Cr O -2+ 2H -------- 3SO -2+ H O + 2Cr +3 2 2 7 4 2 A reaction in a basic solution uses the same method except in a basic solution you must add OH to the H in order to neutralize any acid Electricity is caused by the flow of electrons through a circuit Redox reactions involve the flow of electrons A voltaic cell is an electrochemical cell that produces electrical current from a spontaneous chemical reaction The electrode allows for the flow of electrons through the circuit The salt bridge balances out the positive and negative charges allowing the reaction to continue This reaction occurs because different metals have different orbital energy. This potential is measured in volts joule volt= coulmn The cell potential is a measure of the overall tendency of the redox reaction to occur spontaneously. If the cell potential is positive then the reaction is spontaneous. Oxidation occurs in the anode and reduction occurs in the cathode A shorthand for this concept has been developed: +2 +2 Zn(s)|Zn (aq)||Cu (aq)|Cu(s) This equation can be used to calculate overall cell potential: o o o E Cell Cathode Anode You’ll find that the values of the cathodes and anodes have been calculated already o E cell 0 means that the reaction is spontaneous. The amount of material in the cell has no effect on the cell potential You can calculate the free energy of a cell using this equation: o o ∆ G =−nFE cell N = number of moles F = faraday’s constant (96,485) The Nernst equation calculates cell potential outside of standard conditions: o 0.059v E cell cell logQ n Batteries are voltaic cells that use spontaneous reactions as a source of energy Electrolytic cells are the opposite, their purpose is to induce nonspontaneous reactions by consuming energy

Step 2 of 3

Chapter 40, Problem 36 is Solved
Step 3 of 3

Textbook: Physics for Scientists and Engineers with Modern Physics
Edition: 9
Author: Raymond A. Serway John W. Jewett
ISBN: 9781133954057

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Find the maximum fractional energy loss for a 0.511-MeV