(a) The wavelength and the frequency of the line of the Balmer series for hydrogen. 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 . Balmer interacts with electrons that come from the second energy level (n=2), and Lyman … In which region of electromagnetic spectrum does the Lyman series of hydrogen atom lie (A) Ultraviolet (B) Infra red (C) Visible (D) X-ray. In what region of the electromagnetic spectrum does this series lie ? Hydrogen exhibits several series of line spectra in different spectral regions. Answer Answer: (b) Jump to second orbit leads to Balmer series. Calculate the wavelength of the first line of (i) the Lyman series (ii) the Paschen series. Electrons are falling to the 1-level to produce lines in the Lyman series. The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series tor a hydrogen like ion. Find X assuming R to be same for both H and X? As you notice and the values are decreasing in the series due to the derivation and their state which is Ultra Violet. Calculate

(a) The wavelength and the frequency of the line of the Balmer series for hydrogen. A wavelength of second line of lyman series for H atom is X then wavelength of third line paschen series for the Li2+ ? What hydrogen-like ion has the wavelength difference between the first lines of the Balmer Lyman series equal to ? The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series tor a hydrogen like ion. Just so, is the Lyman series visible? Hope It Helped. 4:04 800+ LIKES. When an electron jumps from the fourth orbit to the second orbit, one gets the (a) second line of Paschen series (b) second line of Balmer series (c) first line of Pfund series (d) second line of Lyman series. Calculate the wavelength of the lowest-energy line in the Lyman series to three significant figures. For example, in the Lyman series, n 1 is always 1. The formation of this line series is due to the ultraviolet emission lines of the hydrogen atom. The wavelength of the first line of Lyman series is 1215 Å, the wavelength of first line of Balmer series will be (A) 4545 Å (B) 5295 Å (C) 6561 Å n 2 is the level being jumped from. As a result the hydrogen like atom 'X' … Balmer series is the spectral series emitted when electron jumps from a higher orbital to orbital of unipositive hydrogen like-species.

(c) Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in LYman series. If $\upsilon_{1}$ is the frequency of the series limit of Lyman series, $\upsilon_{2}$ is the frequency of the first line of Lyman series and $\upsilon_{3}$ is the frequency of the series limit of the Balmer series… A line in the Lyman series of the hydrogen atom emission results from the transition of an electron from the n=3 level to the ground state level. Where, = Wavelength of radiation = Rydberg's Constant = Higher energy level = 3 (second line) = Lower energy level = 1 (Lyman series) Putting the …

(c) Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in LYman series. Chemistry. To what energy level will the hydrogen atom be excited? As En = - 13.6n3 eVAt ground level (n = 1), E1 = - 13.612 = - 13.6 eVAt first excited state (n= 2), E2 = - 13.622 = - 3.4 eVAs hv = E2 - E1 = - 3.4 + 13.6 = 10.2 eV = 1.6 × 10-19 × 10.2 = 1.63 × 10-18 JAlso, c = vλSo λ = cv = chE2 - E1 = (3 x 108) x (6.63 x 10-34)1.63 x 10-18 = 1.22 × 10-7 m ≈ 122 nm A photon of 12.75 eV of energy is absorbed by one electron of a hydrogen atom in the lowest energy level. Find X assuming R H to be same for both H and X? The second line of Lyman series of H coincides with the 6 t h line of Paschen series of an ionic species X. 16.9k VIEWS. Calculate the wavelength of the first line of (i) the Lyman series (ii) the Paschen series. Q.30. Calculate the mass of the deuteron given that the first line in the Lyman series of H lies at 82259.08 cm-1 whereas that of D lies at 82281.476 cm-1. The ratio of energy of the electron in ground state of hydrogen to the electron in first excited state of Be+3 is (a) 4 : 1 (b) 1 : 4 (c) 1 : 8 (d) 8 : 1 20. The ratio of difference in wavelength of 1st and 2nd lines of lyman series in H-like atom to difference in wavelength for 2nd and 3rd line of same series - 6854932 The wavelength of line of the Balmer series is . = Higher energy level = (last line) = Lower energy level = 4 (Bracket series) Putting the values, in above equation, we get. Lyman series is a hydrogen spectral line series that forms when an excited electron comes to the n=1 energy level. 5. Match the correct pairs. 1.6k SHARES. If $\upsilon_{1}$ is the frequency of the series limit of Lyman series, $\upsilon_{2}$ is the frequency of the first line of Lyman series and $\upsilon_{3}$ is the frequency of the series limit of the Balmer series… Calculate the ratio of ionization energies of H and D. Chemistry. Calculate the wavelength of the second line in the Brackett series (nf = 4) of the hydrogen emission spectrum. (a) He+ (b) Li+2 (c) Li+ (d) H 19. For the Balmer series, n 1 is always 2, because electrons are falling to the 2-level. Example \(\PageIndex{1}\): The Lyman Series. In what region of the electromagnetic spectrum does this series lie ? ... the wavelength of the second line of the series should be. calculate the wavelength of the second line in the Brackett series (nf=4) of the hydrogen emission spectrum.

(b) Find the longest and shortest wavelengths in the Lyman series for hydrogen. asked Dec 23, 2018 in Physics by Maryam ( … line indicates transition from 4 --> 2. line indicates transition from 3 -->2. Calculate the wavelength of the second Lyman series and the second line of the Balmer series. (a) (b) (c) (d) H The work function for a metal is 4 eV. The wave length of the first line of the Lyman series of hydrogen is identical to the second line of the Balmer series for some hydrogen like ion 'X'. The second line of Lyman series of H coincides with the 6th line of Paschen series of an Ionic species X. The so-called Lyman series of lines in the emission spectrum of hydrogen corresponds to transitions from various excited states to the n = 1 orbit. The photon radiated from hydrogen corresponding to 2nd line of Lyman series is absorbed by a hydrogen like atom 'X' in 2nd excited state. View Answer Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen. The line spectrum of the Lyman series is formed from transitions of electrons to or from the lowest energy … The second line of Lyman series of H coincides with the sixth line of Paschen series of an ionic species X. Balmer series, the visible region of light, and Lyman series, the UV region of light, each interact with electrons that have ground states in different orbitals. L=4861 = For 3-->2 transition =6562 A⁰

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