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E n = 1 2 m e v 2 k Zq e 2 r n . size 12{E rSub { size 8{n} } = { {1} over {2} } m rSub { size 8{e} } v rSup { size 8{2} } - k { { ital "Zq" rSub { size 8{e} } rSup { size 8{2} } } over {r rSub { size 8{n} } } } } {}

Now we substitute r n size 12{r rSub { size 8{n} } } {} and v size 12{v} {} from earlier equations into the above expression for energy. Algebraic manipulation yields

E n = Z 2 n 2 E 0 ( n = 1, 2, 3, ... ) size 12{E rSub { size 8{n} } = - { {Z rSup { size 8{2} } } over {n rSup { size 8{2} } } } E rSub { size 8{0} } \( n=1," 2, 3, " "." "." "." \) } {}

for the orbital energies of hydrogen-like atoms    . Here, E 0 size 12{E rSub { size 8{0} } } {} is the ground-state energy n = 1 size 12{ left (n=1 right )} {} for hydrogen Z = 1 size 12{ left (Z=1 right )} {} and is given by

E 0 = 2 π 2 q e 4 m e k 2 h 2 = 13.6 eV.

Thus, for hydrogen,

E n = 13.6 eV n 2 ( n = 1, 2, 3, ...).

[link] shows an energy-level diagram for hydrogen that also illustrates how the various spectral series for hydrogen are related to transitions between energy levels.

An energy level diagram is shown. At the left, there is a vertical arrow showing the energy levels increasing from bottom to top. At the bottom, there is a horizontal line showing the energy levels of Lyman series, n is one. The energy is marked as negative thirteen point six electron volt. Then, in the upper half of the figure, another horizontal line showing Balmer series is shown when the value of n is two. The energy level is labeled as negative three point four zero electron volt. Above it there is another horizontal line showing Paschen series. The energy level is marked as negative one point five one electron volt. Above this line, some more lines are shown in a small area to show energy levels of other values of n.
Energy-level diagram for hydrogen showing the Lyman, Balmer, and Paschen series of transitions. The orbital energies are calculated using the above equation, first derived by Bohr.

Electron total energies are negative, since the electron is bound to the nucleus, analogous to being in a hole without enough kinetic energy to escape. As n size 12{n} {} approaches infinity, the total energy becomes zero. This corresponds to a free electron with no kinetic energy, since r n size 12{r rSub { size 8{n} } } {} gets very large for large n size 12{n} {} , and the electric potential energy thus becomes zero. Thus, 13.6 eV is needed to ionize hydrogen (to go from –13.6 eV to 0, or unbound), an experimentally verified number. Given more energy, the electron becomes unbound with some kinetic energy. For example, giving 15.0 eV to an electron in the ground state of hydrogen strips it from the atom and leaves it with 1.4 eV of kinetic energy.

Finally, let us consider the energy of a photon emitted in a downward transition, given by the equation to be

Δ E = hf = E i E f . size 12{ΔE= ital "hf"=E rSub { size 8{i} } - E rSub { size 8{f} } } {}

Substituting E n = ( 13.6 eV / n 2 ) size 12{E rSub { size 8{n} } = - "13" "." 6``"eV"/n rSup { size 8{2} } } {} , we see that

hf = 13.6 eV 1 n f 2 1 n i 2 . size 12{ ital "hf"= left ("13" "." 6" eV" right ) left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}

Dividing both sides of this equation by hc size 12{ ital "hc"} {} gives an expression for 1 / λ size 12{1/λ} {} :

hf hc = f c = 1 λ = 13.6 eV hc 1 n f 2 1 n i 2 . size 12{ { { ital "hf"} over { ital "hc"} } = { {f} over {c} } = { {1} over {λ} } = { { left ("13" "." 6" eV" right )} over { ital "hc"} } left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}

It can be shown that

13.6 eV hc = 13.6 eV 1.602 × 10 −19 J/eV 6.626 × 10 −34 J·s 2.998 × 10 8 m/s = 1.097 × 10 7 m –1 = R

is the Rydberg constant    . Thus, we have used Bohr’s assumptions to derive the formula first proposed by Balmer years earlier as a recipe to fit experimental data.

1 λ = R 1 n f 2 1 n i 2 size 12{ { {1} over {λ} } =R left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}

We see that Bohr’s theory of the hydrogen atom answers the question as to why this previously known formula describes the hydrogen spectrum. It is because the energy levels are proportional to 1 / n 2 size 12{1/n rSup { size 8{2} } } {} , where n size 12{n} {} is a non-negative integer. A downward transition releases energy, and so n i size 12{n rSub { size 8{i} } } {} must be greater than n f size 12{n rSub { size 8{f} } } {} . The various series are those where the transitions end on a certain level. For the Lyman series, n f = 1 size 12{n rSub { size 8{f} } =1} {} — that is, all the transitions end in the ground state (see also [link] ). For the Balmer series, n f = 2 size 12{n rSub { size 8{f} } =2} {} , or all the transitions end in the first excited state; and so on. What was once a recipe is now based in physics, and something new is emerging—angular momentum is quantized.

Triumphs and limits of the bohr theory

Bohr did what no one had been able to do before. Not only did he explain the spectrum of hydrogen, he correctly calculated the size of the atom from basic physics. Some of his ideas are broadly applicable. Electron orbital energies are quantized in all atoms and molecules. Angular momentum is quantized. The electrons do not spiral into the nucleus, as expected classically (accelerated charges radiate, so that the electron orbits classically would decay quickly, and the electrons would sit on the nucleus—matter would collapse). These are major triumphs.

Questions & Answers

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
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Friday Reply
what is vector quantity
Ridwan Reply
Vector quality have both direction and magnitude, such as Force, displacement, acceleration and etc.
Besmellah
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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
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Salaudeen
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je
mutual induction can be define as the current flowing in one coil that induces a voltage in an adjacent coil.
Johnson
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Ajayi Reply
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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
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Mavis
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Ademiye
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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
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Emmanuel Reply
what is a matter
Yinka
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Yinka
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Yinka
what is a matter
Yinka
I want the nuclear physics conversation
Mohamed
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Clara
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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