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An electron is in the nth Bohr orbit of the hydrogen atom.

(a) Show that the period of the electron is

 

and determine the numerical value of to.

(b) On average, an electron remains in the n=2 orbit for approximately 10 microseconds before jump down to n= 1 orbit. How many revolutions does the electron make in the excited state?

(c) Define the period of one revolution as a one-electron year, analogous to an earth year is the period of the earth's motion around the sun. Explain whether we should think of the electron in n=2 orbit "as living for a long time"

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Vikas
VikasLv10
10 Mar 2021

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

An electron of momentum p is at a distance r from a stationary proton. The electron has kinetic energy . The atoms has potential energy and total energy . If the electron is bound to the proton to form a hydrogen atom, its average position is at the proton but the uncertainty in its position is approximately equal to the radius r of its orbit. The electron's average vector momentum is zero, but its average squared momentum is approximately equal to the squared uncertainty in its momentum as given by the uncertainty principle. Treating the atom as a one-dimensional system: (a) estimate the uncertainty in the electron's momentum in terms of r. Estimate the electron's (b) kinetic energy and (c) total energy in terms of r. The actual value of r is the one that minimizes the total energy, resulting in a stable atom. Find (d) that value of r and (e) the resulting total energy. (f) State how your answers compare with the predictions of the Bohr theory.

calculate the angular momentum of the earth in its orbit around the sun?

 

 

m   5.97 10 kg 6 E R   6.38 10 m Orbital radius 11 r   1.50 10 m Period of rotation Prot = 24 h 86,400 s  Period of revolution Prev = 7 1 y 3.156 10 s   Identify: L I z z   Set Up: For a particle of mass m moving in a circular path at a distance r from the axis, 2 I mr  and v r   . For a uniform sphere of mass M and radius R and an axis through its center, 2 2 5 I MR  . The earth has mass 24 E m   5.97 10 kg , radius 6 E R   6.38 10 m and orbit radius 11 r   1.50 10 m . The earth completes one rotation on its axis in 24 h 86,400 s  and one orbit in 7 1 y 3.156 10 s   . Execute: (a) 2 24 11 2 40 2 7 2 rad (5.97 10 kg)(1.50 10 m) 2.67 10 kg m /s 3.156 10 s L I mr z z z                   . The radius of the earth is much less than its orbit radius, so it is very reasonable to model it as a particle for this calculation. (b)   2 2 2 24 6 2 33 2 5 5 2 rad (5.97 10 kg)(6.38 10 m) 7.07 10 kg m /s 86,400 s L I MR z z                  Evaluate: The angular momentum associated with each of these motions is very large. Note that the orbital angular momentum is over 7 orders of magnitude greater than its rotational angular momentum. In fact, most of the total angular momentum of the solar system comes from the sum of the orbital angular momenta of its planets.

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