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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

Why is the sky blue...?
Star Reply
It's filtered light from the 2 forms of radiation emitted from the sun. It's mainly filtered UV rays. There's a theory titled Scatter Theory that covers this topic
Mike
A heating coil of resistance 30π is connected to a 240v supply for 5min to boil a quantity of water in a vessel of heat capacity 200jk. If the initial temperature of water is 20°c and it specific heat capacity is 4200jkgk calculate the mass of water in a vessel
fasawe Reply
A thin equi convex lens is placed on a horizontal plane mirror and a pin held 20 cm vertically above the lens concise in position with its own image the space between the undersurface of d lens and the mirror is filled with water (refractive index =1•33)and then to concise with d image d pin has to
Azummiri Reply
Be raised until its distance from d lens is 27cm find d radius of curvature
Azummiri
what happens when a nuclear bomb and atom bomb bomb explode add the same time near each other
FlAsH Reply
A monkey throws a coconut straight upwards from a coconut tree with a velocity of 10 ms-1. The coconut tree is 30 m high. Calculate the maximum height of the coconut from the top of the coconut tree? Can someone answer my question
Fatinizzah Reply
v2 =u2 - 2gh 02 =10x10 - 2x9.8xh h = 100 ÷ 19.6 answer = 30 - h.
Ramonyai
why is the north side is always referring to n side of magnetic
sam Reply
who is a nurse
Chilekwa Reply
A nurse is a person who takes care of the sick
Bukola
a nurse is also like an assistant to the doctor
Gadjawa
explain me wheatstone bridge
Malik Reply
good app
samuel
Wheatstone bridge is an instrument used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component.
MUHD
Rockwell Software is Rockwell Automation’s "Retro Encabulator". Now, basically the only new principle involved is that instead of power being generated by the relative motion of conductors and fluxes, it’s produced by the modial interaction of magneto-reluctance and capacitive diractance. The origin
Chip
what refractive index
Adjah Reply
write a comprehensive note on primary colours
Harrison Reply
relationship between refractive index, angle of minimum deviation and angle of prism
Harrison
Who knows the formula for binding energy,and what each variable or notation stands for?
Agina Reply
1. A black thermocouple measures the temperature in the chamber with black walls.if the air around the thermocouple is 200 C,the walls are at 1000 C,and the heat transfer constant is 15.compute the temperature gradient
Tikiso Reply
what is the relationship between G and g
Olaiya Reply
G is the u. constant, as g stands for grav, accelerate at a discreet point
Mark
Is that all about it?
Olaiya
pls explain in details
Olaiya
G is a universal constant
Mark
g stands for the gravitational acceleration point. hope this helps you.
Mark
balloon TD is at a gravitational acceleration at a specific point
Mark
I'm sorry this doesn't take dictation very well.
Mark
Can anyone explain the Hooke's law of elasticity?
Olaiya Reply
extension of a spring is proportional to the force applied so long as the force applied does not exceed the springs capacity according to my textbook
Amber
does this help?
Amber
Yes, thanks
Olaiya
so any solid can be compressed how compressed is dependent upon how much force is applied F=deltaL
Amber
sorry, the equation is F=KdeltaL delta is the triangle symbol and L is length so the change in length is proportional to amount of Force applied I believe that is what Hookes law means. anyone catch any mistakes here please correct me :)
Amber
I think it is used only for solids and not liquids, isn't it?
Olaiya
basically as long as you dont exceed the elastic limit the object should return to it original form but if you exceed this limit the object will not return to original shape as it will break
Amber
Thanks for the explanation
Olaiya
yh, liquids don't apply here, that should be viscosity
Chiamaka
hope it helps 😅
Amber
also, an object doesnt have to break necessarily, but it will have a new form :)
Amber
Yes
Olaiya
yeah, I think it is for solids but maybe there is a variation for liquids? that I am not sure of
Amber
ok
Olaiya
good luck!
Amber
Same
Olaiya
aplease i need a help on spcific latent heat of vibrations
Bilgate
specific latent heat of vaporisation
Bilgate
how many kilometers makes a mile
Margaret Reply
about 1.6 kilometres.
Faizyab
near about 1.67 kilometers
Aakash
equal to 1.609344 kilometers.
MUHD
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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