Explanation The principal quantum number which is denoted by n can take any integer values ( 1 , 2 , 3 , ...... ) . The orbital quantum number which is denoted by l can take any integer values from 0 up to n − 1 . The magnetic quantum number which is denoted by m l can take any integer values from − l to + l . The spin quantum number which is denoted by m s can be + 1 2 or − 1 2 . Write the expression to obtain the number states possible. N = ∑ l = 0 l = n − 1 2 ( 2 l + 1 ) (I) Here, N is the total number of possible states, n is the principal quantum number and l is the orbital quantum number. The quantum number of each state is given as, ( n , l , m l , m s ) . Conclusion: Thus, for n = 5 , the possible values of l are, l = 0 , 1 , 2 , 3 , 4 . Thus, for l = 4 , the possible values of m l are, m l = − 4 , − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 , 4 . The values m s can only be, m s = + 1 2 , − 1 2 Substitute 5 for n in equation (I) to calculate N . N = 2 ( 2 ( 0 ) + 1 ) + 2 ( 2 ( 1 ) + 1 ) + 2 ( 2 ( 2 ) + 1 ) + 2 ( 2 ( 3 ) + 1 ) + 2 ( 2 ( 4 ) + 1 ) = 2 ( 1 ) + 2 ( 3 ) + 2 ( 5 ) + 2 ( 7 ) + 2 ( 9 ) = 2 + 6 + 10 + 14 + 18 = 50 All the possible quantum number are listed below; ( 5 , 0 , 0 , − 1 2 ) , ( 5 , 0 , 0 , + 1 2 ) , ( 5 , 1 , 1 , − 1 2 ) , ( 5 , 1 , 1 , + 1 2 ) , ( 5 , 1 , 0 , − 1 2 ) , ( 5 , 1 , 0 , + 1 2 ) , ( 5 , 1 , − 1 , − 1 2 ) , ( 5 , 1 , − 1 , + 1 2 ) , ( 5 , 2 , 2 , − 1 2 ) , ( 5 , 2 , 2 , + 1 2 ) , ( 5 , 2 , 1 , − 1 2 ) , ( 5 , 2 , 1 , + 1 2 ) , ( 5 , 2 , 0 , − 1 2 ) , ( 5 , 2 ,

Physics for Scientists and Engineers with Modern Physics

4th Edition

ISBN: 9780131495081

Author: Douglas C. Giancoli

Publisher: Addison-Wesley

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Chapter 39, Problem 3P

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The number of possible different states possible for an electron whose principal quantum number is n=5 and the values of quantum number of each state.

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Chapter 39, Problem 2P

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How many electron states are there in the following shells: (a) n = 4, (b) n = 1, (c) n = 3, (d) n = 2?

(i) Using Bohr model for atomic hydrogen, obtain energy levels for the 2s, 3s and 3p states in the actual number with the unit of [eV]. We consider a transition that electron in the 3p state emits a photon and make a transition to the 2s state. What is the frequency v of this photon ? (ii) Now we do not include electron spin angular momentum, and just estimate an effect of a magnetic field B on this transition (Normal Zeeman effect) with orbital angular momentum. How many lines of optical transition do we expect ? What is the interval of the frequency in the field B = 0.1 Tesla ? (iii) In this situation, we do not expect transition from 3s to 2s state if the electron is initially in the 3s state, Explain the reason. (iv) We now consider an effect of magnetic field B to a free electron spin (not in Hydrogen, but a free electron). The magnetic field of B = 1.0 Tesla will split the energy level into two (Zeeman) levels. Obtain the level difference in the unit of [eV] from the value of…

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Physics for Scientists and Engineers with Modern Physics

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The number of possible different states possible for an electron whose principal quantum number is n = 5 and the values of quantum number of each state.

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The number of possible different states possible for an electron whose principal quantum number is n=5 and the values of quantum number of each state.

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Chapter 39 Solutions