One of the lines in the balmer series of the hydrogen atom emission spectrum is at 397nm.?
Q. One of the lines in the balmer series of the hydrogen atom emission spectrum is at 397nm. It results from a transition from an upper energy level to n=2. What is the principal quantum number of the upper level?
Asked by right_512 - Sun Apr 26 23:14:20 2009 - - 1 Answers - 0 Comments
A. 1/y = (n^-2 - n'^-2)R where R is rydberg constant n= initial state n' final state R = 1.0973 * 10^7 solving we get n' = 7
Answered by not 'relative' only abso - Mon Apr 27 05:01:41 2009
Q. One of the lines in the balmer series of the hydrogen atom emission spectrum is at 397nm. It results from a transition from an upper energy level to n=2. What is the principal quantum number of the upper level?
Asked by right_512 - Sun Apr 26 23:14:20 2009 - - 1 Answers - 0 Comments
A. 1/y = (n^-2 - n'^-2)R where R is rydberg constant n= initial state n' final state R = 1.0973 * 10^7 solving we get n' = 7
Answered by not 'relative' only abso - Mon Apr 27 05:01:41 2009
What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lymann...?
Q. What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lymann series?
Asked by melissa - Sat Apr 11 00:41:21 2009 - - 1 Answers - 0 Comments
A. 1/L=R(1/m^2 - 1/n^2) L=wavelength For minimum wavelength, take n = infinity So 1/L=R/m^2 i.e. L=m^2/R For Lyman m=1 and for Balmer m=2 L_Balmer/L_Lyman = 2^2/1^2 =4 Ratio is 4
Answered by unknown - Sat Apr 11 13:48:39 2009
Q. What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lymann series?
Asked by melissa - Sat Apr 11 00:41:21 2009 - - 1 Answers - 0 Comments
A. 1/L=R(1/m^2 - 1/n^2) L=wavelength For minimum wavelength, take n = infinity So 1/L=R/m^2 i.e. L=m^2/R For Lyman m=1 and for Balmer m=2 L_Balmer/L_Lyman = 2^2/1^2 =4 Ratio is 4
Answered by unknown - Sat Apr 11 13:48:39 2009
What is the longest possible wavelength for a line in the Balmer series? What is the shortest?
Q. I'm trying to do my Chem homework but I cannot seem to find the answer to this one anywhere, please help me.
Asked by Shawn N - Tue Oct 2 11:30:14 2007 - - 1 Answers - 1 Comments
A. The Balmer series is calculated using the Balmer formula, an empirical equation discovered by Johann Balmer in 1885. The visible spectrum of light from hydrogen displays four wavelengths, 410 nm, 434 nm, 486 nm, and 656 nm, that reflect emissions of photons by electrons in excited states transitioning to the quantum level described by the principal quantum number n equals 2. So the shortest is 410 nm and longest is 656 nm. I hope it helps
Answered by Ehsan R - Tue Oct 2 11:58:41 2007
Q. I'm trying to do my Chem homework but I cannot seem to find the answer to this one anywhere, please help me.
Asked by Shawn N - Tue Oct 2 11:30:14 2007 - - 1 Answers - 1 Comments
A. The Balmer series is calculated using the Balmer formula, an empirical equation discovered by Johann Balmer in 1885. The visible spectrum of light from hydrogen displays four wavelengths, 410 nm, 434 nm, 486 nm, and 656 nm, that reflect emissions of photons by electrons in excited states transitioning to the quantum level described by the principal quantum number n equals 2. So the shortest is 410 nm and longest is 656 nm. I hope it helps
Answered by Ehsan R - Tue Oct 2 11:58:41 2007
Balmer series calculating the 4 longest wavelenghts?
Q. I need help with this question: 1.Calculate the wavelengths in nanometers of the four lines of the Balmer series with the longest wavelengths. Attempt: I know the formula 1/ = 1/(2^2) - 1/(m^2) My question is: what are the values of m for the 4 longest wavelenghts? Is there a maximum value of m? 2. What is the wavelength of the line with the shortest wavelength in the Balmer series??
Asked by Bob C - Thu May 28 08:24:40 2009 - - 2 Answers - 0 Comments
A. Your formula is missing the Rydberg constant multiplying the right hand side. RH = 10,973,731.57 m^-1 (for an infinitely heavy H-atom nucleus). After taking into account the finite relative mass of the H-atom nucleus (relative to an electron, that is), the number to use is 0.010972 nm^1: 1/lambda = (0.010972)(1/2^2 - 1/m^2) nm^1 The emission in the Balmer series is to the n = 2 level from higher n values (n = principal quantum no. for the H-atom). The longest wavelengths are the lowest in energy, so the emission is from the level closest in energy to the n = 2 level. The 4 longest wavelengths therefore correspond to m = 3, 4, 5, and 6 in your formula. Theoretically, there is no maximum for m. The Hydrogen atom energy levels get… [cont.]
Answered by unknown - Thu May 28 08:48:39 2009
Q. I need help with this question: 1.Calculate the wavelengths in nanometers of the four lines of the Balmer series with the longest wavelengths. Attempt: I know the formula 1/ = 1/(2^2) - 1/(m^2) My question is: what are the values of m for the 4 longest wavelenghts? Is there a maximum value of m? 2. What is the wavelength of the line with the shortest wavelength in the Balmer series??
Asked by Bob C - Thu May 28 08:24:40 2009 - - 2 Answers - 0 Comments
A. Your formula is missing the Rydberg constant multiplying the right hand side. RH = 10,973,731.57 m^-1 (for an infinitely heavy H-atom nucleus). After taking into account the finite relative mass of the H-atom nucleus (relative to an electron, that is), the number to use is 0.010972 nm^1: 1/lambda = (0.010972)(1/2^2 - 1/m^2) nm^1 The emission in the Balmer series is to the n = 2 level from higher n values (n = principal quantum no. for the H-atom). The longest wavelengths are the lowest in energy, so the emission is from the level closest in energy to the n = 2 level. The 4 longest wavelengths therefore correspond to m = 3, 4, 5, and 6 in your formula. Theoretically, there is no maximum for m. The Hydrogen atom energy levels get… [cont.]
Answered by unknown - Thu May 28 08:48:39 2009
What is the shortest wavelength in the Balmer series (n=6)?
Q. Basically this homework question is asking me to calculate the shortest wavelength of the radiation emitted by the hydrogen atom when a transition occurs from level n = 6. I have tried to use -Rhc x (8/36) and then divide the speed of light by that answer but I guess I'm completely wrong. Can anyone help?
Asked by mschroeder24 - Sun Jul 6 16:10:29 2008 - - 3 Answers - 0 Comments
A. When a hydrogen electron transitions from level 6 to level 2 (level 2 is the lower limit in the Balmer series) the wavelength of radiation (w) can be found by the Rydberg equation: 1/w = R(1/4 - 1/36), where 1/4 and 1/36 are the reciprocals of the squares of the lower and upper energy levels (6 & 2) respectively, and R is the Rydberg constant, which has been experimentally determined as 1,096,776 per meter. The value in parentheses (1/4 - 1/36) is equal to 0.222 and multiplying this by R gives 243,728 per meter. Take the reciprocal of that value and we find the wavelength w = 4.103 * 10**-7 meter, or 410.3 nanometers. This is a line in the deep violet end of the visible spectrum. Hope this helps.
Answered by Reginald - Sun Jul 6 18:01:04 2008
Q. Basically this homework question is asking me to calculate the shortest wavelength of the radiation emitted by the hydrogen atom when a transition occurs from level n = 6. I have tried to use -Rhc x (8/36) and then divide the speed of light by that answer but I guess I'm completely wrong. Can anyone help?
Asked by mschroeder24 - Sun Jul 6 16:10:29 2008 - - 3 Answers - 0 Comments
A. When a hydrogen electron transitions from level 6 to level 2 (level 2 is the lower limit in the Balmer series) the wavelength of radiation (w) can be found by the Rydberg equation: 1/w = R(1/4 - 1/36), where 1/4 and 1/36 are the reciprocals of the squares of the lower and upper energy levels (6 & 2) respectively, and R is the Rydberg constant, which has been experimentally determined as 1,096,776 per meter. The value in parentheses (1/4 - 1/36) is equal to 0.222 and multiplying this by R gives 243,728 per meter. Take the reciprocal of that value and we find the wavelength w = 4.103 * 10**-7 meter, or 410.3 nanometers. This is a line in the deep violet end of the visible spectrum. Hope this helps.
Answered by Reginald - Sun Jul 6 18:01:04 2008
What is the frequency of the limiting line in the Balmer series?
Q. What is the frequency of the limiting line in the Balmer series?
Asked by Shyam Sundar - Sat Jun 3 03:36:46 2006 - - 2 Answers - 0 Comments
A. Hi visit the site below..It should help..
Answered by MONJIT - Tue Jun 6 06:53:03 2006
Q. What is the frequency of the limiting line in the Balmer series?
Asked by Shyam Sundar - Sat Jun 3 03:36:46 2006 - - 2 Answers - 0 Comments
A. Hi visit the site below..It should help..
Answered by MONJIT - Tue Jun 6 06:53:03 2006
What is the highest frequenc of radiation in Balmer Series of the hydrogen atom?
Q. Mass of proton is 938.28 MeV/c^2 Mass of electron is 0.511 MeV/c^2
Asked by ntchmst - Mon May 12 13:26:45 2008 - - 2 Answers - 0 Comments
A. 410.2 nm
Answered by rijim2001 - Mon May 12 13:32:46 2008
Q. Mass of proton is 938.28 MeV/c^2 Mass of electron is 0.511 MeV/c^2
Asked by ntchmst - Mon May 12 13:26:45 2008 - - 2 Answers - 0 Comments
A. 410.2 nm
Answered by rijim2001 - Mon May 12 13:32:46 2008
What is the wavelength of the least energetic photon emitted in Balmer Series of the Hydrogen atom?
Q. Mass of electron is 0.511 MeV/c^2 Mass of proton is 938.28 MeV/c^2
Asked by ntchmst - Mon May 12 13:19:06 2008 - - 2 Answers - 0 Comments
A. I assume you are supposed to show our work as to how you calculated it. The lower frequency is the Balmer alpha line, which is important in astronomy and is 656nm But you can work out for yourself how to get it...
Answered by Stardustspeck - Mon May 12 13:30:04 2008
Q. Mass of electron is 0.511 MeV/c^2 Mass of proton is 938.28 MeV/c^2
Asked by ntchmst - Mon May 12 13:19:06 2008 - - 2 Answers - 0 Comments
A. I assume you are supposed to show our work as to how you calculated it. The lower frequency is the Balmer alpha line, which is important in astronomy and is 656nm But you can work out for yourself how to get it...
Answered by Stardustspeck - Mon May 12 13:30:04 2008
Electron termination Balmer Series?
Q. I understand why electrons go up energy levels for example from n=2 to n=4 and then they fall back down and emit light, my question is why do electrons terminate at a certain level. What makes it stop at that energy level for example why do all electrons in a Balmer series terminate at n=2 why cant they terminate at n=5 or n=6 ?
Asked by Andres - Tue Jul 7 10:34:55 2009 - - 2 Answers - 0 Comments
A. It has to do with the history of who studied what when. The names for the various series are: Lyman (n=1), Balmer, Paschen, Brachett and Pfund. Balmer (n=2) was first because those lines are in the visible and near visible region of the spectrum.
Answered by paul1 - Tue Jul 7 11:10:24 2009
Q. I understand why electrons go up energy levels for example from n=2 to n=4 and then they fall back down and emit light, my question is why do electrons terminate at a certain level. What makes it stop at that energy level for example why do all electrons in a Balmer series terminate at n=2 why cant they terminate at n=5 or n=6 ?
Asked by Andres - Tue Jul 7 10:34:55 2009 - - 2 Answers - 0 Comments
A. It has to do with the history of who studied what when. The names for the various series are: Lyman (n=1), Balmer, Paschen, Brachett and Pfund. Balmer (n=2) was first because those lines are in the visible and near visible region of the spectrum.
Answered by paul1 - Tue Jul 7 11:10:24 2009
one of the lines of the balmer series of the hydrogen emission...?
Q. one of the lines of the balmer series of the hydrogen emission spectrum is at 397 nm. what is the principle quantum number for the original energy level for this transition. show all steps in solving this problem Could someone please help.. I'm not looking for an answer but just a guide. I'm more into how to do this than the answer... my class just learned this stuff today and it's due tomorrow..
Asked by Mandy - Tue Nov 7 18:25:48 2006 - - 1 Answers - 0 Comments
A. This is probably too late for your class, but... Find the energy of each energy level. (for hydrogen, E1 = 1312 kJ/mol; E of any other level = En = 1312/n^2) Find the amount of energy (in kJ/mol) of the 397 nm line. (E = hf, f = c/ lambda, don't forget to use Avogadro's number) Match the energy level to the energy of the light. Good luck !
Answered by wibblytums - Wed Nov 8 15:31:00 2006
Q. one of the lines of the balmer series of the hydrogen emission spectrum is at 397 nm. what is the principle quantum number for the original energy level for this transition. show all steps in solving this problem Could someone please help.. I'm not looking for an answer but just a guide. I'm more into how to do this than the answer... my class just learned this stuff today and it's due tomorrow..
Asked by Mandy - Tue Nov 7 18:25:48 2006 - - 1 Answers - 0 Comments
A. This is probably too late for your class, but... Find the energy of each energy level. (for hydrogen, E1 = 1312 kJ/mol; E of any other level = En = 1312/n^2) Find the amount of energy (in kJ/mol) of the 397 nm line. (E = hf, f = c/ lambda, don't forget to use Avogadro's number) Match the energy level to the energy of the light. Good luck !
Answered by wibblytums - Wed Nov 8 15:31:00 2006
What is the shortest wavelength in the Balmer series (n=2)?
Q. a) 486.17 nm b) 656.34 nm c) 364.63 nm Also, can you please explain the anwser. Thanks.
Asked by lullabyfreedom - Wed Dec 6 16:00:32 2006 - - 1 Answers - 0 Comments
A. the answer is b. use the equation frequency=3.2881x10^15(1/ 4-1/n^2). 3.2881x10^15 is a constant its units are 1/s. n is always a whole number. n also has to be greater than 2. when u plug in n=2 to the equation you get... frequency=4.5668x10^14 then use c=vy where c is the speed of light v is frequency and y is wavelength. you should get 6.5634x10^-7 or 656.34 nm
Answered by Katie B - Wed Dec 6 16:39:32 2006
Q. a) 486.17 nm b) 656.34 nm c) 364.63 nm Also, can you please explain the anwser. Thanks.
Asked by lullabyfreedom - Wed Dec 6 16:00:32 2006 - - 1 Answers - 0 Comments
A. the answer is b. use the equation frequency=3.2881x10^15(1/ 4-1/n^2). 3.2881x10^15 is a constant its units are 1/s. n is always a whole number. n also has to be greater than 2. when u plug in n=2 to the equation you get... frequency=4.5668x10^14 then use c=vy where c is the speed of light v is frequency and y is wavelength. you should get 6.5634x10^-7 or 656.34 nm
Answered by Katie B - Wed Dec 6 16:39:32 2006
Bohr Model and the Balmer Series?
Q. The question says... Calculate th energy of each Bohr orbit from n=2 to n=7 in kJ/mole. E= -1312 kJ / n^2 mole i dont understand what this question is asking. what does the n stand for? is there something i need to plug in? Thanks for your help!
Asked by c - Sat Jan 12 20:50:35 2008 - - 2 Answers - 0 Comments
A. What they want you to calculate is the 7th energy level and the 2nd energy level. In a hydrogen atom (Bhor model), when an electron is excited to the 7th level and it drops to the 2nd level a photo of 397.0nm (violet) wavelength is produced. So all you have to do is plug in 2 and 7 for in the equation you are given to find the energy of each orbit. E = -1312kJ/n E(7) = -1312kJ/49 = -26.775kJ/mole E(2) = -1312kJ/4 = -328kJ/mole It is the difference in these two energies that produces that aforementioned photon of violet light.
Answered by cat_lover - Sat Jan 12 20:59:36 2008
Q. The question says... Calculate th energy of each Bohr orbit from n=2 to n=7 in kJ/mole. E= -1312 kJ / n^2 mole i dont understand what this question is asking. what does the n stand for? is there something i need to plug in? Thanks for your help!
Asked by c - Sat Jan 12 20:50:35 2008 - - 2 Answers - 0 Comments
A. What they want you to calculate is the 7th energy level and the 2nd energy level. In a hydrogen atom (Bhor model), when an electron is excited to the 7th level and it drops to the 2nd level a photo of 397.0nm (violet) wavelength is produced. So all you have to do is plug in 2 and 7 for in the equation you are given to find the energy of each orbit. E = -1312kJ/n E(7) = -1312kJ/49 = -26.775kJ/mole E(2) = -1312kJ/4 = -328kJ/mole It is the difference in these two energies that produces that aforementioned photon of violet light.
Answered by cat_lover - Sat Jan 12 20:59:36 2008
why would any line in the hydrogen spectrum between 250nm and 700nm belong to the Balmer series?
Q. why would any line in the hydrogen spectrum between 250nm and 700nm belong to the Balmer series?
Asked by Master - Sat Dec 30 14:08:20 2006 - - 2 Answers - 0 Comments
A. Because Balmer is the one who discovered it Wavelenght=Blamer constant * (m^2/(m^2-n^2)) n=2 and m> 2 Balmer constant= 364.56nm
Answered by Suhas - Sat Dec 30 14:17:22 2006
Q. why would any line in the hydrogen spectrum between 250nm and 700nm belong to the Balmer series?
Asked by Master - Sat Dec 30 14:08:20 2006 - - 2 Answers - 0 Comments
A. Because Balmer is the one who discovered it Wavelenght=Blamer constant * (m^2/(m^2-n^2)) n=2 and m> 2 Balmer constant= 364.56nm
Answered by Suhas - Sat Dec 30 14:17:22 2006
Why is Balmer series...?
Q. the way it is? R(1/ 2^2 - 1/n^2) the general equation is R( 1/ Ninitial^2 - 1/ Nfinal^2) in this case we r dropping to N =2, so Nfinal = 2 so y is 2 in the left of the 2 fractions?
Asked by thebigred - Tue Sep 29 01:38:01 2009 - - 1 Answers - 0 Comments
A. Al's answer covers the essentials of the Bohr model and the Balmer series. But I think your question refers more specifically to the order of the denominator terms. The general equation needs to be sign-adjusted for a positive result regardless of the direction of the electron jump, since the result is the inverse wavelength 1/ of the absorbed or emitted photon. Without this adjustment 1/ would come out negative for all emitted photons (due to downward jumps where 1/Ninitial^2<1/Nfinal^2). Either the sign of R is adjusted, or the order of terms can be swapped with the same result. EDIT: Or, to paraphrase Al, ignore the order and just take the absolute value so the wavelength is positive.
Answered by kirchwey - Thu Oct 1 14:03:18 2009
Q. the way it is? R(1/ 2^2 - 1/n^2) the general equation is R( 1/ Ninitial^2 - 1/ Nfinal^2) in this case we r dropping to N =2, so Nfinal = 2 so y is 2 in the left of the 2 fractions?
Asked by thebigred - Tue Sep 29 01:38:01 2009 - - 1 Answers - 0 Comments
A. Al's answer covers the essentials of the Bohr model and the Balmer series. But I think your question refers more specifically to the order of the denominator terms. The general equation needs to be sign-adjusted for a positive result regardless of the direction of the electron jump, since the result is the inverse wavelength 1/ of the absorbed or emitted photon. Without this adjustment 1/ would come out negative for all emitted photons (due to downward jumps where 1/Ninitial^2<1/Nfinal^2). Either the sign of R is adjusted, or the order of terms can be swapped with the same result. EDIT: Or, to paraphrase Al, ignore the order and just take the absolute value so the wavelength is positive.
Answered by kirchwey - Thu Oct 1 14:03:18 2009
what are the different series (Lyman,Balmer,Brakket and P-fund ,Pascal ) and what are their uses?
Q. I want detail discribtion of these series and what are the uses of these sereis,meanswhat we can predict by this series/
Asked by shashi l - Sun Oct 1 00:24:50 2006 - - 2 Answers - 0 Comments
A. the spectrum obtained from hydrogen gas consists of series of lines called line spectrum of H2.They are, 1.Lyman series : seen in U.V region. 2.Balmer series : seen in visible region 3. Paschen series:seen in Infrared region 4.Bracket series:seen in Infrared region 5.Pfund series:seen in Infrared region The line spectrum of an element is the charecteristic of that element and is different from other element.
Answered by malukollam20 - Sun Oct 1 02:22:00 2006
Q. I want detail discribtion of these series and what are the uses of these sereis,meanswhat we can predict by this series/
Asked by shashi l - Sun Oct 1 00:24:50 2006 - - 2 Answers - 0 Comments
A. the spectrum obtained from hydrogen gas consists of series of lines called line spectrum of H2.They are, 1.Lyman series : seen in U.V region. 2.Balmer series : seen in visible region 3. Paschen series:seen in Infrared region 4.Bracket series:seen in Infrared region 5.Pfund series:seen in Infrared region The line spectrum of an element is the charecteristic of that element and is different from other element.
Answered by malukollam20 - Sun Oct 1 02:22:00 2006
in each transition, an amount of energy is emitted that is equal to the difference?
Q. in each transition, an amount of energy is emitted that is equal to the difference in energy between level n=2 and the excited state from which the electron fell. These transitions generate the Balmer series of bright lines. Which transition can be related to the violet line in the Balmer spectrum? Which transition can be related to the red line? Explain please and ill give u the best answer.
Asked by lil21dog - Wed Apr 5 20:56:53 2006 - - 1 Answers - 0 Comments
A. i presume you have the energy values for the different levels? if so you use the equation E=hc/lamda to find the wavelength for each energy transition in the series. The red line will be about 575nm and the purple line will be about 375nm.
Answered by Scott and Jay - Sat Apr 8 10:15:11 2006
Q. in each transition, an amount of energy is emitted that is equal to the difference in energy between level n=2 and the excited state from which the electron fell. These transitions generate the Balmer series of bright lines. Which transition can be related to the violet line in the Balmer spectrum? Which transition can be related to the red line? Explain please and ill give u the best answer.
Asked by lil21dog - Wed Apr 5 20:56:53 2006 - - 1 Answers - 0 Comments
A. i presume you have the energy values for the different levels? if so you use the equation E=hc/lamda to find the wavelength for each energy transition in the series. The red line will be about 575nm and the purple line will be about 375nm.
Answered by Scott and Jay - Sat Apr 8 10:15:11 2006
Calculate the energy of a photon?
Q. For a hydrogen atom, calculate the energy of a photon in the Balmer series that results from the transition n=3 to n=2. Answer should be in J.
Asked by sgrajeda05 - Fri Oct 31 15:30:49 2008 - - 1 Answers - 0 Comments
Q. For a hydrogen atom, calculate the energy of a photon in the Balmer series that results from the transition n=3 to n=2. Answer should be in J.
Asked by sgrajeda05 - Fri Oct 31 15:30:49 2008 - - 1 Answers - 0 Comments
WAVELENGTH of line in the BRACKET series?!?
Q. Hydrogen exhibits several series of line spectral regions. For example, the Lyman series (nf=1 in Balmer-Rydberg equation) occurs in ultraviolet region while the balmer (nf=2) series occurs in the visible range and the Paschen (nf=3), Brackett (nf=4) and Pfund (nf=5) series all occur in the infrared range. What is the wavelength (in nm) of a lin ein the bracket series where ni=7?
Asked by strawberryminiwheats - Wed Nov 5 17:57:17 2008 - - 1 Answers - 0 Comments
A. The equation to figure this out is 1/wavelength = R(1/nf^2-1/ni^2) with R being Rydberg constant. Then you would solve for wavelength.
Answered by Laura W - Wed Nov 5 20:14:05 2008
Q. Hydrogen exhibits several series of line spectral regions. For example, the Lyman series (nf=1 in Balmer-Rydberg equation) occurs in ultraviolet region while the balmer (nf=2) series occurs in the visible range and the Paschen (nf=3), Brackett (nf=4) and Pfund (nf=5) series all occur in the infrared range. What is the wavelength (in nm) of a lin ein the bracket series where ni=7?
Asked by strawberryminiwheats - Wed Nov 5 17:57:17 2008 - - 1 Answers - 0 Comments
A. The equation to figure this out is 1/wavelength = R(1/nf^2-1/ni^2) with R being Rydberg constant. Then you would solve for wavelength.
Answered by Laura W - Wed Nov 5 20:14:05 2008
which of these sentences are true?
Q. Which of the following statements are true? Pick all that apply. A. Two ways for atoms to emit photons is by thermal and electrical excitation. B. The wavelength of light can be measured by a diffraction grating. C. The visible lines of the hydrogen spectrum are members of Balmer series. D. Electrons generally occupy the state with the highest possible energy. E. Atoms of different elements have the same energy levels and therefore elemental composition cannot be determined by spectroscopic analysis. F. When electrons jump from a higher energy level to a lower energy level, photons are emitted.
Asked by curious - Thu Apr 16 16:50:34 2009 - - 1 Answers - 0 Comments
A. a b f
Answered by unknown - Thu Apr 16 16:57:27 2009
Q. Which of the following statements are true? Pick all that apply. A. Two ways for atoms to emit photons is by thermal and electrical excitation. B. The wavelength of light can be measured by a diffraction grating. C. The visible lines of the hydrogen spectrum are members of Balmer series. D. Electrons generally occupy the state with the highest possible energy. E. Atoms of different elements have the same energy levels and therefore elemental composition cannot be determined by spectroscopic analysis. F. When electrons jump from a higher energy level to a lower energy level, photons are emitted.
Asked by curious - Thu Apr 16 16:50:34 2009 - - 1 Answers - 0 Comments
A. a b f
Answered by unknown - Thu Apr 16 16:57:27 2009
Which of the following series of Hydrogen is primarily ultraviolet?
Q. A Lyman B Balmer C Paschen D Brackett E Pfund
Asked by Mrgoodhair - Wed Oct 22 14:59:12 2008 - - 1 Answers - 0 Comments
A. A. Lyman See:
Answered by Chem Man - Wed Oct 22 15:03:47 2008
Q. A Lyman B Balmer C Paschen D Brackett E Pfund
Asked by Mrgoodhair - Wed Oct 22 14:59:12 2008 - - 1 Answers - 0 Comments
A. A. Lyman See:
Answered by Chem Man - Wed Oct 22 15:03:47 2008
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