For an excited state of hydrogen, the smallest angle the orbital angular momentum vector L → can have with the z -axis is ( θ L ) min = arccos ( n − 1 n ( n − 1 ) ) .

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Answer 1

Question

Chapter 41, Problem 41.53P

(a)

To determine

To show: For an excited state of hydrogen, the smallest angle the orbital angular momentum vector L→ can have with the z -axis is

(θL)min=arccos(n−1n(n−1)) .

(b)

To determine

The largest possible angle, (θL)max between L→ and the z -axis.

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(a) Calculate: (i) the energy spacing AE between the ground state and the first excited state of the hydrogen atom; (ii) and the ratio AE/E between the spacing and the ground state energy. (b) Consider now a macroscopic system: a simple pendulum which consists of a 5 g mass attached to a 2 m long, massless and inextensible string. Calculate (i) the total energy E1 of the pendulum when the string makes an angle of 60° with the vertical; (ii) the frequeney of the pendulum's small oscillations and the energy AE of one quantum; and (iii) the ratio AE/E1. (c) Examine the sizes of the ratio AE/E1 calculated in parts (a) and (b) and comment on the importance of the quantum effects for the hydrogen atom and the pendulum.

(a) What is the magnitude of the orbital angular momentum in a state with e = 2? (b) What is the magnitude of its largest projection on an imposed axis? (a) Number 2.50998008 Units J.s (b) Number 2.11 Units J.s

a) How many distinct angles from the vertical axis can the orbital angular momentum vector L make for an electron with l = 7? b)Calculate the smallest possible angle the L can make with respect to the vertical axis. (Hint: The smallest angle occurs when ml takes the maximum allowed value. Sketch L in that case and compare the vertical component, which is related to ml, to the magnitude of L, which is related to l.)

Answer 2

Question

Chapter 41, Problem 41.53P

(a)

To determine

To show: For an excited state of hydrogen, the smallest angle the orbital angular momentum vector L→ can have with the z -axis is

(θL)min=arccos(n−1n(n−1)) .

(b)

To determine

The largest possible angle, (θL)max between L→ and the z -axis.

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 41, Problem 41.53P

(a)

To determine

To show: For an excited state of hydrogen, the smallest angle the orbital angular momentum vector L→ can have with the z -axis is

(θL)min=arccos(n−1n(n−1)) .

(b)

To determine

The largest possible angle, (θL)max between L→ and the z -axis.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 41, Problem 41.53P

(a)

To determine

To show: For an excited state of hydrogen, the smallest angle the orbital angular momentum vector L→ can have with the z -axis is

(θL)min=arccos(n−1n(n−1)) .

(b)

To determine

The largest possible angle, (θL)max between L→ and the z -axis.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 41, Problem 41.53P

(a)

To determine

To show: For an excited state of hydrogen, the smallest angle the orbital angular momentum vector L→ can have with the z -axis is

(θL)min=arccos(n−1n(n−1)) .

(b)

To determine

The largest possible angle, (θL)max between L→ and the z -axis.

Answer 6

Chapter 41, Problem 41.53P

(a)

To determine

To show: For an excited state of hydrogen, the smallest angle the orbital angular momentum vector L→ can have with the z -axis is

(θL)min=arccos(n−1n(n−1)) .

Answer 7

Chapter 41, Problem 41.53P

(a)

To determine

To show: For an excited state of hydrogen, the smallest angle the orbital angular momentum vector L→ can have with the z -axis is

(θL)min=arccos(n−1n(n−1)) .

Answer 8

(b)

To determine

The largest possible angle, (θL)max between L→ and the z -axis.

Answer 9

(b)

To determine

The largest possible angle, (θL)max between L→ and the z -axis.

Answer 10

Expert Solution & Answer

Answer 11

Expert Solution & Answer

Answer 12

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Answer 13

Ch. 41.1 - Prob. 41.1TYU Ch. 41.2 - Prob. 41.2TYU Ch. 41.3 - Prob. 41.3TYU Ch. 41.4 - In this section we assumed that the magnetic field...Ch. 41.5 - In which of the following situations is the...Ch. 41.6 - Prob. 41.6TYU Ch. 41.7 - Prob. 41.7TYU Ch. 41.8 - Prob. 41.8TYU Ch. 41 - Prob. 41.1DQ Ch. 41 - Prob. 41.2DQ

For an excited state of hydrogen, the smallest angle the orbital angular momentum vector L → can have with the z -axis is ( θ L ) min = arccos ( n − 1 n ( n − 1 ) ) .

Solution Summary: The author explains how the orbital angular momentum vector stackreltoL can have a smallest angle with the z -axis, and the equation for the

To show: For an excited state of hydrogen, the smallest angle the orbital angular momentum vector L→ can have with the z -axis is

(θL)min=arccos(n−1n(n−1)) .

The largest possible angle, (θL)max between L→ and the z -axis.

Want to see the full answer?

Chapter 41 Solutions

For an excited state of hydrogen, the smallest angle the orbital angular momentum vector L → can have with the z -axis is ( θ L ) min = arccos ( n − 1 n ( n − 1 ) ) .

Solution Summary: The author explains how the orbital angular momentum vector stackreltoL can have a smallest angle with the z -axis, and the equation for the

To show: For an excited state of hydrogen, the smallest angle the orbital angular momentum vector L→ can have with the z -axis is (θL)min=arccos(n−1n(n−1)) .

The largest possible angle, (θL)max between L→ and the z -axis.

Want to see the full answer?

Chapter 41 Solutions

To show: For an excited state of hydrogen, the smallest angle the orbital angular momentum vector L→ can have with the z -axis is

(θL)min=arccos(n−1n(n−1)) .