Question

The transition from n = 6 to n = 2 produces the Balmer series line with...

The transition from n = 6 to n = 2 produces the Balmer series line with wavelength λ0. The wavelength is 4 times larger than an n = x to n = 1 transition in the Lyman series. Determine the value of x.

Group of answer choices

3

5

6

2

4

Homework Answers

Answer #1

For Balmer series we can write the wavelength as,

Substituting values we can write,

For Layman series we can write,

Now, in the case of layman series the value of n is x and the wavelength is 1/4 times the wavelength in the case of Balmer series. Hence we can write,

Dividing the above two equations for Balmer and Layman series we can write,

Therefore the correct answer is (a).

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
a) Determine the wavelength of the second Balmer line (n = 4 to n = 2...
a) Determine the wavelength of the second Balmer line (n = 4 to n = 2 transition) using Figure A. b) Determine the wavelength of the third Lyman line. c) Determine the wavelength of the first Balmer line Figure A
1) The last two visible line observed in the Balmer series are of the wavelength 434.0...
1) The last two visible line observed in the Balmer series are of the wavelength 434.0 nm and 410.2 nm. For each of the wavelength, a) Determine the frequency associated with the photon associated with this wavelength. b) Determine the amount of energy released by this photon. c) The Balmer series are the results of excited electron at higher level relaxing to the electronic level of principle quantum number of nf = 2. Determine the excited electronic level, ni, of...
1) The aqua color line of the Balmer series occurs at 486.1 nm, a) Determine the...
1) The aqua color line of the Balmer series occurs at 486.1 nm, a) Determine the frequency associated with the photon associated with this wavelength b) Determine the amount of energy released by this photon. c) The Balmer series are the results of excited electron at higher level relaxing to the electronic level of principle quantum number of nf = 2. Determine the excited electronic level, ni, of this electron. d) If this excited electron has the same wavelength as...
Hydrogen exhibits several series of line spectra in different spectral regions. For example the Lyman series...
Hydrogen exhibits several series of line spectra in different spectral regions. For example the Lyman series (nf = 1 in Balmer-Rydberg equation) occurs in the 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 shortest-wavelength (in nm) in the Brackett series?
The visible emission lines observed by Balmer all involved nf = 2. Which of the following...
The visible emission lines observed by Balmer all involved nf = 2. Which of the following is the best explanation of why the lines with nf = 3 are not observed in the visible portion of the spectrum? Transitions to nf = 3 emit photons in the ultraviolet portion of the spectrum. Transitions to nf = 3 emit photons in the infrared portion of the spectrum. Transitions to nf = 3 emit photons that are at exactly the same wavelengths...
For a hydrogen atom, calculate the wavelength of an emitted photon in the Lyman series that...
For a hydrogen atom, calculate the wavelength of an emitted photon in the Lyman series that results from the transition n = 3 to n = 1. The Rydberg constant is 2.18 x 10^-18 J.
Astronomers use the line emission from the quantum state n = 3 to n = 2...
Astronomers use the line emission from the quantum state n = 3 to n = 2 to probe ionized hydrogen gas – that is, hydrogen gas stripped of its electron. In ionization equilibrium, a balance is established between the process of ionization and its reverse, recombination, in which a free electron e − and proton p recombine to form neutral atomic hydrogen H, releasing one or more photons γ : e − + p ↔ H + γ The recombination...
1. Calculate the energy (in kJ/mol) of the Lyman series line for hydrogen that is, the...
1. Calculate the energy (in kJ/mol) of the Lyman series line for hydrogen that is, the transition from n=2 to n=1 (1 kJ/mol= 8.6 cm^-1) 2. Would you expect the energy of the same transition as in the first problem to be greater or less for the helium ion, He^+? Why? Make a sketch to relative scale the energy levels that give rise to the Lyman bands in H-atoms and He^+.
The Kα line (transition from n=2 to n=1) in an unknown metal sample is observed at...
The Kα line (transition from n=2 to n=1) in an unknown metal sample is observed at a wavelength of 0.0205 nm. Which metal is it? 27Co (Cobalt) 78Pt (Platinum) 94Pu (Plutonium) 47Ag (Silver)
The Balmer lines are from a) The transition of hydrogen atoms into the lowest energy level...
The Balmer lines are from a) The transition of hydrogen atoms into the lowest energy level b) The transition of hydrogen atoms into the n=2 level. c) The transition of helium atoms into the lowest energy level. d) The transition of oxygen atoms to the lowest energy level. Emission lines are caused when an electron moves ____ energy; Absorption lines are caused when an electron moves ______ energy a) to higher; to lower b) to a higher n; to a...