<< Chapter < Page Chapter >> Page >

Explain how Bohr’s rule for the quantization of electron orbital angular momentum differs from the actual rule.

What is a hydrogen-like atom, and how are the energies and radii of its electron orbits related to those in hydrogen?

Problems&Exercises

By calculating its wavelength, show that the first line in the Lyman series is UV radiation.

1 λ = R 1 n f 2 1 n i 2 λ = 1 R ( n i n f ) 2 n i 2 n f 2 ; n i = 2, n f = 1, 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 ) drarrow λ= { {1} over {R} } left [ { { \( n rSub { size 8{i} } cdot n rSub { size 8{f} } \) rSup { size 8{2} } } over {n rSub { size 8{i} } rSup { size 8{2} } - n rSub { size 8{f} } rSup { size 8{2} } } } right ]; n rSub { size 8{i} } =2" , "n rSub { size 8{f} } =1" ,"} {} so that

λ = m 1.097 × 10 7 ( 2 × 1 ) 2 2 2 1 2 = 1 . 22 × 10 7 m = 122 nm size 12{λ= left ( { {m} over {1 "." "097" times "10" rSup { size 8{7} } } } right ) left [ { { \( 2 "." 1 \) rSup { size 8{2} } } over {4 - 1} } right ]=1 "." "22" times "10" rSup { size 8{ - 7} } " m"="122"" nm"} {} , which is UV radiation.

Find the wavelength of the third line in the Lyman series, and identify the type of EM radiation.

Look up the values of the quantities in a B = h 2 4 π 2 m e kq e 2 , and verify that the Bohr radius a B is 0.529 × 10 10 m size 12{0 "." "529" times "10" rSup { size 8{ - "10"} } " m"} {} .

a B = h 2 4 π 2 m e kZq e 2 = ( 6.626 × 10 34 J·s ) 2 4 π 2 ( 9.109 × 10 31 kg ) ( 8.988 × 10 9 N · m 2 / C 2 ) ( 1 ) ( 1.602 × 10 19 C ) 2 = 0.529 × 10 10 m

Verify that the ground state energy E 0 size 12{E rSub { size 8{0} } } {} is 13.6 eV by using E 0 = 2 π 2 q e 4 m e k 2 h 2 . size 12{E rSub { size 8{o} } = { {2π rSup { size 8{2} } q rSub { size 8{e} } rSup { size 8{4} } m rSub { size 8{e} } k rSup { size 8{2} } } over {h rSup { size 8{2} } } } "." } {}

If a hydrogen atom has its electron in the n = 4 size 12{n=4} {} state, how much energy in eV is needed to ionize it?

0.850 eV

A hydrogen atom in an excited state can be ionized with less energy than when it is in its ground state. What is n size 12{n} {} for a hydrogen atom if 0.850 eV of energy can ionize it?

Find the radius of a hydrogen atom in the n = 2 size 12{n=2} {} state according to Bohr’s theory.

2.12 × 10 –10 m size 12{2 "." "12" times "10" rSup { size 8{"-10"} } " m"} {}

Show that 13.6 eV / hc = 1.097 × 10 7 m = R (Rydberg’s constant), as discussed in the text.

What is the smallest-wavelength line in the Balmer series? Is it in the visible part of the spectrum?

365 nm

It is in the ultraviolet.

Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest wavelength in the series.

Do the Balmer and Lyman series overlap? To answer this, calculate the shortest-wavelength Balmer line and the longest-wavelength Lyman line.

No overlap

365 nm

122 nm

(a) Which line in the Balmer series is the first one in the UV part of the spectrum?

(b) How many Balmer series lines are in the visible part of the spectrum?

(c) How many are in the UV?

A wavelength of 4 . 653 μm size 12{4 "." "653 μm"} {} is observed in a hydrogen spectrum for a transition that ends in the n f = 5 size 12{n rSub { size 8{f} } =5} {} level. What was n i size 12{n rSub { size 8{i} } } {} for the initial level of the electron?

7

A singly ionized helium ion has only one electron and is denoted He + size 12{"He" rSup { size 8{+{}} } } {} . What is the ion’s radius in the ground state compared to the Bohr radius of hydrogen atom?

A beryllium ion with a single electron (denoted Be 3 + size 12{"Be" rSup { size 8{3+{}} } } {} ) is in an excited state with radius the same as that of the ground state of hydrogen.

(a) What is n size 12{n} {} for the Be 3 + size 12{"Be" rSup { size 8{3+{}} } } {} ion?

(b) How much energy in eV is needed to ionize the ion from this excited state?

(a) 2

(b) 54.4 eV

Atoms can be ionized by thermal collisions, such as at the high temperatures found in the solar corona. One such ion is C + 5 size 12{C rSup { size 8{+5} } } {} , a carbon atom with only a single electron.

(a) By what factor are the energies of its hydrogen-like levels greater than those of hydrogen?

(b) What is the wavelength of the first line in this ion’s Paschen series?

(c) What type of EM radiation is this?

Verify Equations r n = n 2 Z a B and a B = h 2 4 π 2 m e kq e 2 = 0.529 × 10 10 m using the approach stated in the text. That is, equate the Coulomb and centripetal forces and then insert an expression for velocity from the condition for angular momentum quantization.

kZq e 2 r n 2 = m e V 2 r n , so that r n = kZq e 2 m e V 2 = kZq e 2 m e 1 V 2 . From the equation m e vr n = n h 2 π , we can substitute for the velocity, giving: r n = kZq e 2 m e 4 π 2 m e 2 r n 2 n 2 h 2 so that r n = n 2 Z h 2 4 π 2 m e kq e 2 = n 2 Z a B , where a B = h 2 2 m e kq e 2 .

The wavelength of the four Balmer series lines for hydrogen are found to be 410.3, 434.2, 486.3, and 656.5 nm. What average percentage difference is found between these wavelength numbers and those predicted by 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 )} {} ? It is amazing how well a simple formula (disconnected originally from theory) could duplicate this phenomenon.

Questions & Answers

how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!
QuizOver.com Reply
Practice Key Terms 7

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, College physics -- hlca 1104. OpenStax CNX. May 18, 2013 Download for free at http://legacy.cnx.org/content/col11525/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics -- hlca 1104' conversation and receive update notifications?

Ask