Concept explainers
(a)
The radius of the orbit of the hydrogen atom in the first excited state.
(a)
Answer to Problem 41P
The radius of the orbit of the hydrogen atom in the first excited state is
Explanation of Solution
Write the expression for the radius of the orbit in the hydrogen atom relating the Bohr radius.
Here,
Conclusion:
Substitute
Therefore, the radius of the orbit of the hydrogen atom in the first excited state is
(b)
The linear momentum of the electron in the hydrogen atom.
(b)
Answer to Problem 41P
The linear momentum of the electron in the hydrogen atom is
Explanation of Solution
Write the expression for the columbic force of attraction.
Here,
Write the expression for the
Here,
Equate equation (II) and (III) to solve for
Write the expression for the linear momentum.
Here,
Use equation (IV) to solve for
Conclusion:
Substitute
Therefore, The linear momentum of the electron in the hydrogen atom is
(c)
The
(c)
Answer to Problem 41P
The angular momentum of the electron is
Explanation of Solution
Write the expression for the angular momentum.
Here,
Conclusion:
Substitute
Therefore, the angular momentum of the electron is
(d)
The kinetic energy of the electron.
(d)
Answer to Problem 41P
The kinetic energy of the electron is
Explanation of Solution
Write the expression for the kinetic energy of the electron.
Use equation (IV) to solve for
Conclusion:
Substitute
Therefore, the kinetic energy of the electron is
(e)
The potential energy of the electron.
(e)
Answer to Problem 41P
The potential energy of the electron is
Explanation of Solution
Write the expression for the potential energy.
Here,
Conclusion:
Substitute
Therefore, the potential energy of the electron is
(f)
The total energy of the system.
(f)
Answer to Problem 41P
The total energy of the system is
Explanation of Solution
Write the expression for the total energy of the system.
Here,
Conclusion:
Substitute
Therefore, the total energy of the system is
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Chapter 11 Solutions
Principles of Physics
- A star with mass M and radius R collides head-on with another star of mass ¾*M and radius 4/5*R, and they coalesce to form a new start at rest whose radius is 6/5*R. Assume that initially the colliding stars had angular velocities with opposite directions but the same magnitude w. What is the magnitude and direction of the final’s stars angular velocity? (Express the magnitude as a fraction of w.)arrow_forwardUsing partial derivatives, calculate the propagated uncertainty in the mass in the following case: given the centripetal force Fc = (20.0 ± 0.5) N, the angular velocity w = (29.2 ± 0.3) rad/s, and the radius R = (0.12 ± 0.01) m get the mass value,m = Fc / (w2R). Express the result in the form m = m + Δm ----------------------------------- THAT'S THE QUESTION ASKED, see the image for the answer. Also have a look at the second image, the blue one. --------------------------------------------------------------- Explain what is the 1/m just after the equals sign at the second line of the answer. Also, explain why the answer does not use the square root just like the blue image, of if it is using it. Then, say in which case should I use the partial derivate to calculate the uncertainty.arrow_forwardA merry-go-round is a playground ride that consists of a large disk mounted to that it can freely rotate in a horizontal plane. The merry-go-round shown is initially at rest, has a radius R = 1.3 meters, and a mass M = 291 kg. A small boy of mass m = 42 kg runs tangentially to the merry-go-round at a speed of v = 1.8 m/s, and jumps on. Randomized VariablesR = 1.3 metersM = 291 kgm = 42 kgv = 1.8 m/s Part A- Calculate the moment of inertia of the merry-go-round, in kg ⋅ m2. Part B- Immediately before the boy jumps on the merry go round, calculate his angular speed (in radians/second) about the central axis of the merry-go-round. Part C- Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy. Part D- The boy then crawls towards the center of the merry-go-round along a radius. What is the angular speed in radians/second of the merry-go-round when the boy is half way between the edge and the center of the merry…arrow_forward
- Answer without rounding off: (COLLAB Gauss's Law for Mass) Journey through the Center of the Earth. A 1024-kg blue ball is dropped from an initial z-position of 2.1 x 106 m through the center of a planet with a radius of 7.7 x 106 m. If the mass of the planet is 46.5 x 1015 kg, measure the displacement of the ball at time t = 6 s?arrow_forwardA star has initially a radius of 780000000 m and a period of rotation about its axis of 22 days. Eventually it changes into a neutron star with a radius of only 25000 m and a period of 0.1 s. Assuming that the mass has not changed, find Assume a star has the shape of a sphere. (Suggestion: do it with formula first, then put the numbers in) [Recommended time : 5-8 minutes] (a) the ratio of initial to final angular momentum (Li/Lf) a. 1.85E+16 b. 51.2 c. 0.0195 d. 5.4E-17 (b) the ratio of initial to final kinetic energy a. 2.84E-24 b. 371000 c. 2.69E-6 d. 3.52E+23arrow_forwardA merry-go-round is a playground ride that consists of a large disk mounted to that it can freely rotate in a horizontal plane. The merry-go-round shown is initially at rest, has a radius R = 1.3 meters, and a mass M = 251 kg. A small boy of mass m = 42 kg runs tangentially to the merry-go-round at a speed of v = 1.2 m/s, and jumps on. Randomized Variables R = 1.3 metersM = 251 kgm = 42 kgv = 1.2 m/s a)Calculate the moment of inertia of the merry-go-round, in kg ⋅ m2. b) Immediately before the boy jumps on the merry go round, calculate his angular speed (in radians/second) about the central axis of the merry-go-round. c) Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy.arrow_forward
- A merry-go-round is a playground ride that consists of a large disk mounted to that it can freely rotate in a horizontal plane. The merry-go-round shown is initially at rest, has a radius R = 1.3 meters, and a mass M = 251 kg. A small boy of mass m = 42 kg runs tangentially to the merry-go-round at a speed of v = 1.2 m/s, and jumps on. Randomized Variables R = 1.3 metersM = 251 kgm = 42 kgv = 1.2 m/s a) The boy then crawls towards the center of the merry-go-round along a radius. What is the angular speed in radians/second of the merry-go-round when the boy is half way between the edge and the center of the merry go round? b) The boy then crawls to the center of the merry-go-round. What is the angular speed in radians/second of the merry-go-round when the boy is at the center of the merry go round? c)Finally, the boy decides that he has had enough fun. He decides to crawl to the outer edge of the merry-go-round and jump off. Somehow, he manages to jump in such a way that he hits the…arrow_forwardA merry-go-round is a playground ride that consists of a large disk mounted to that it can freely rotate in a horizontal plane. The merry-go-round shown is initially at rest, has a radius R = 1.5 meters, and a mass M = 251 kg. A small boy of mass m = 41 kg runs tangentially to the merry-go-round at a speed of v = 1.8 m/s, and jumps on.Randomized VariablesR = 1.5 metersM = 251 kgm = 41 kgv = 1.8 m/s (a) Calculate the moment of inertia of the merry-go-round, in kg ⋅ m2. Part (b) Immediately before the boy jumps on the merry go round, calculate his angular speed (in radians/second) about the central axis of the merry-go-round.(c) Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy.(d) The boy then crawls towards the center of the merry-go-round along a radius. What is the angular speed in radians/second of the merry-go-round when the boy is half way between the edge and the center of the merry go round?(e) The…arrow_forwardAn asteroid has a mass of m = 2.6 x 106 kg and is approaching Earth. When the asteroid is exactly 3 radii away from the Earth's centre, it's speed relative to the Earth's centre is u = 8.7 x 103 ms-1. The asteroid then falls to the Earth's surface, but remains intact without dissipating any energy as it passes through the Earth's atmosphere. If the rotation of the asteroid and the Earth is ignored, what is the kinetic energy of the asteroid just before it hits the ground? The Earth has mass ME = 5.98 x 1024 kg and a radius of 6.38 x 106 m. To find the relevant potential energies, you must use G = 6.67 x 10-11 N m2 kg-2arrow_forward
- The figure shows a spherical hollow inside a lead sphere of radius R = 4.3 m; the surface of the hollow passes through the center of the sphere and “touches” the right side of the sphere. The mass of the sphere before hollowing was M = 388 kg. With what gravitational force does the hollowed-out lead sphere attract a small sphere of mass m = 37 kg that lies at a distance d = 12 m from the center of the lead sphere, on the straight line connecting the centers of the spheres and of the hollow?arrow_forwardYou plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the earth with enough speed to make it to the moon. Some information that will help during this problem: mearth = 5.9742 x 1024 kgrearth = 6.3781 x 106 mmmoon = 7.36 x 1022 kgrmoon = 1.7374 x 106 mdearth to moon = 3.844 x 108 m (center to center)G = 6.67428 x 10-11 N-m2/kg2 1) On your first attempt you leave the surface of the earth at v = 5534 m/s. How far from the center of the earth will you get? 2) Since that is not far enough, you consult a friend who calculates (correctly) the minimum speed needed as vmin = 11068 m/s. If you leave the surface of the earth at this speed, how fast will you be moving at the surface of the moon? Hint carefully write out an expression for the potential and kinetic energy of the ship on the surface of earth, and on the surface of moon. Be sure to include the gravitational potential energy of the earth even when the ship is…arrow_forward(Gauss's Law for Mass) Journey through the Center of the Earth. A 1024-kg blue ball is dropped from an initial z-position of 5.8 x 106 m through the center of a planet with radius 7.1 x 106 m. If the mass of the planet is 41.5 x 1015 kg, measure the displacement of the ball at time t = 4 s?arrow_forward
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