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L = m e vr n = n h 2 π n = 1, 2, 3, , size 12{ left (n=1,2,3, dotslow right )} {}

where L is the angular momentum, m e is the electron’s mass, r n is the radius of the n th orbit, and h is Planck’s constant. Note that angular momentum is L = . For a small object at a radius r , I = mr 2 and ω = v / r , so that L = mr 2 v / r = mvr . Quantization says that this value of mvr can only be equal to h / 2, 2 h / 2, 3 h / 2 size 12{h/2,` 2h/2, `3h/2} {} , etc. At the time, Bohr himself did not know why angular momentum should be quantized, but using this assumption he was able to calculate the energies in the hydrogen spectrum, something no one else had done at the time.

From Bohr’s assumptions, we will now derive a number of important properties of the hydrogen atom from the classical physics we have covered in the text. We start by noting the centripetal force causing the electron to follow a circular path is supplied by the Coulomb force. To be more general, we note that this analysis is valid for any single-electron atom. So, if a nucleus has Z size 12{Z} {} protons ( Z = 1 size 12{Z=1} {} for hydrogen, 2 for helium, etc.) and only one electron, that atom is called a hydrogen-like atom    . The spectra of hydrogen-like ions are similar to hydrogen, but shifted to higher energy by the greater attractive force between the electron and nucleus. The magnitude of the centripetal force is m e v 2 / r n size 12{m rSub { size 8{e} } v rSup { size 8{2} } /r rSub { size 8{n} } } {} , while the Coulomb force is k Zq e q e / r n 2 size 12{k left ( ital "Zq" rSub { size 8{e} } right ) left (q rSub { size 8{e} } right )/r rSub { size 8{n} } rSup { size 8{2} } } {} . The tacit assumption here is that the nucleus is more massive than the stationary electron, and the electron orbits about it. This is consistent with the planetary model of the atom. Equating these,

k Zq e 2 r n 2 = m e v 2 r n (Coulomb = centripetal). size 12{k { { ital "Zq" rSub { size 8{e} } rSup { size 8{2} } } over {r rSub { size 8{n} } rSup { size 8{2} } } } = { {m rSub { size 8{e} } v rSup { size 8{2} } } over {r rSub { size 8{n} } } } } {}

Angular momentum quantization is stated in an earlier equation. We solve that equation for v size 12{v} {} , substitute it into the above, and rearrange the expression to obtain the radius of the orbit. This yields:

r n = n 2 Z a B , for allowed orbits size 12{r rSub { size 8{n} } = { {n rSup { size 8{2} } } over {Z} } a rSub { size 8{B} } } {} n = 1, 2, 3, , size 12{ left (n=1, 2, 3, dotslow right )} {}

where a B size 12{a rSub { size 8{B} } } {} is defined to be the Bohr radius    , since for the lowest orbit n = 1 size 12{ left (n=1 right )} {} and for hydrogen Z = 1 size 12{ left (Z=1 right )} {} , r 1 = a B size 12{r rSub { size 8{1} } =a rSub { size 8{B} } } {} . It is left for this chapter’s Problems and Exercises to show that the Bohr radius is

a B = h 2 2 m e kq e 2 = 0.529 × 10 10 m . size 12{a rSub { size 8{B} } = { {h rSup { size 8{2} } } over {4π rSup { size 8{2} } m rSub { size 8{e} } ital "kq" rSub { size 8{e} } rSup { size 8{2} } } } =0 "." "529" times "10" rSup { size 8{ - "10"} } " m"} {}

These last two equations can be used to calculate the radii of the allowed (quantized) electron orbits in any hydrogen-like atom . It is impressive that the formula gives the correct size of hydrogen, which is measured experimentally to be very close to the Bohr radius. The earlier equation also tells us that the orbital radius is proportional to n 2 size 12{n rSup { size 8{2} } } {} , as illustrated in [link] .

The electron orbits are shown in the form of four concentric circles. The radius of each circle is marked as r sub one, r sub two, up to r sub four.
The allowed electron orbits in hydrogen have the radii shown. These radii were first calculated by Bohr and are given by the equation r n = n 2 Z a B size 12{r rSub { size 8{n} } = { {n rSup { size 8{2} } } over {Z} } a rSub { size 8{B} } } {} . The lowest orbit has the experimentally verified diameter of a hydrogen atom.

To get the electron orbital energies, we start by noting that the electron energy is the sum of its kinetic and potential energy:

E n = KE + PE . size 12{E rSub { size 8{n} } =" KE "+" PE"} {}

Kinetic energy is the familiar KE = 1 / 2 m e v 2 size 12{ ital "KE"= left (1/2 right )m rSub { size 8{e} } v rSup { size 8{2} } } {} , assuming the electron is not moving at relativistic speeds. Potential energy for the electron is electrical, or PE = q e V size 12{ ital "PE"=q rSub { size 8{e} } V} {} , where V size 12{V} {} is the potential due to the nucleus, which looks like a point charge. The nucleus has a positive charge Zq e size 12{ ital "Zq" rSub { size 8{e} } } {} ; thus, V = kZq e / r n , recalling an earlier equation for the potential due to a point charge. Since the electron’s charge is negative, we see that PE = kZq e / r n size 12{ ital "PE"= - ital "kZq" rSub { size 8{e/r rSub { size 6{n} } } } } {} . Entering the expressions for KE size 12{ ital "KE"} {} and PE size 12{ ital "PE"} {} , we find

Questions & Answers

the meaning of phrase in physics
Chovwe Reply
is the meaning of phrase in physics
Chovwe
write an expression for a plane progressive wave moving from left to right along x axis and having amplitude 0.02m, frequency of 650Hz and speed if 680ms-¹
Gabriel Reply
how does a model differ from a theory
Friday Reply
what is vector quantity
Ridwan Reply
Vector quality have both direction and magnitude, such as Force, displacement, acceleration and etc.
Besmellah
Is the force attractive or repulsive between the hot and neutral lines hung from power poles? Why?
Jack Reply
what's electromagnetic induction
Chinaza Reply
electromagnetic induction is a process in which conductor is put in a particular position and magnetic field keeps varying.
Lukman
wow great
Salaudeen
what is mutual induction?
je
mutual induction can be define as the current flowing in one coil that induces a voltage in an adjacent coil.
Johnson
how to undergo polarization
Ajayi Reply
show that a particle moving under the influence of an attractive force mu/y³ towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v²k² and distance uk²/√u-vk as origin
Gabriel Reply
show that a particle moving under the influence of an attractive force mu/y^3 towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v^2k^2 and distance uk^2/√u-k as origin
Gabriel Reply
No idea.... Are you even sure this question exist?
Mavis
I can't even understand the question
Ademiye
yes it was an assignment question "^"represent raise to power pls
Gabriel
mu/y³ u>v²k² uk²/√u-vk please help me out
Gabriel
An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10kg . Pendulum 2 has a bob with a mass of 100 kg . Describe how the motion of the pendula will differ if the bobs are both displaced by 12º .
Imtiaz Reply
no ideas
Augstine
if u at an angle of 12 degrees their period will be same so as their velocity, that means they both move simultaneously since both both hovers at same length meaning they have the same length
Ademiye
Modern cars are made of materials that make them collapsible upon collision. Explain using physics concept (Force and impulse), how these car designs help with the safety of passengers.
Isaac Reply
calculate the force due to surface tension required to support a column liquid in a capillary tube 5mm. If the capillary tube is dipped into a beaker of water
Mildred Reply
find the time required for a train Half a Kilometre long to cross a bridge almost kilometre long racing at 100km/h
Ademiye
method of polarization
Ajayi
What is atomic number?
Makperr Reply
The number of protons in the nucleus of an atom
Deborah
type of thermodynamics
Yinka Reply
oxygen gas contained in a ccylinder of volume has a temp of 300k and pressure 2.5×10Nm
Taheer Reply
Practice Key Terms 7

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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