Share Question. The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen-like ion. Download the PDF Question Papers Free for off line practice and view the Solutions online. The spectral lines are grouped into series according to n′. Then which of the following is correct? The atomic number ‘Z’ of hydrogen like ion is _____ And, this energy level is the lowest energy level of the hydrogen atom. Paiye sabhi sawalon ka Video solution sirf photo khinch kar. The wavelength of the first line of Lyman series of hydrogen atom is equal to that of the second line of Balmer series of a hydrogen like ion . For example, the 2 → 1 line is called "Lyman-alpha" (Ly-α), while the 7 → 3 line is called "Paschen-delta” (Pa-δ). Where, = Wavelength of radiation = Rydberg's Constant = Higher energy level = 2 = Lower energy level = 1 (Lyman series) Putting the values, in above equation, we get Thus . The wavelengths in the hydrogen spectrum with m=1 form a series of spectral lines called the Lyman series. The wavelength of first line of Balmer series is 6563Å. The wavelength of the first line of Lyman series of hydrogen is 1216 A. Zigya App. Brackett of the United States and Friedrich Paschen of Germany. Related Questions: Copy Link. 3.4k SHARES. Create. This formula gives a wavelength of lines in the Lyman series of the hydrogen spectrum. The Lyman series of the Hydrogen Spectral Emissions is the first level where n' = 1. Rutherfords experiment on scattering of particles showed for the first time that the atom has (a) electrons (b) protons (c) nucleus (d) neutrons Lyman series is a hydrogen spectral line series that forms when an excited electron comes to the n=1 energy level. The wavelength of the first line of Lyman series in hydrogen atom is 1216. Be the first to write the explanation for this question by commenting below. Calculate the wavelengths of the first four members of the Lyman series i… Add To Playlist Add to Existing Playlist. Q. 2. The wavelength of the second line of the same series will be. 678.4 Å Find the ratio of series limit wavelength of Balmer series to wavelength of first time line of paschen series. The photon liberated a photoelectron from a stationary H atom in ground state. Lyman series is obtained when an electron jumps from n>1 to n = 1 energy level of hydrogen atom. Option A is correct. The first line in the Lyman series in the spectrum of hydrogen atom occurs at a wavelength of 1215 Å and the limit for Balmer series is 3645 Å. For example, in the Lyman series, n 1 is always 1. 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 R = Rydberg constant = 1.097 × 10 +7 m. n 1 = 1 n 2 = 2. … Can you explain this answer? Calculate the wavelength of the first line in the Lyman series and show that… 02:05. a. Let v 1 be the frequency of series limit of Lyman series, v 2 the frequency of the first line of Lyman series, and v 3 the frequency of series limit of Balmer series. α line of Lyman series p = 1 and n = 2; α line of Lyman series p = 1 and n = 3; γ line of Lyman series p = 1 and n = 4; the longest line of Lyman series p = 1 and n = 2; the shortest line of Lyman series p = 1 and n = ∞ Different lines of Lyman series are . Maximum wave length corresponds to minimum frequency i.e., n 1 = 1, n 2 = 2. 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. Assuming f to be frequency of first line in Balmer series, the frequency of the immediate next( ie, second) line is a) 0.50 / b)1.35 / c)2.05 / d)2.70 / Be the first to write the explanation for this question by commenting below. Options (a) 2/9 λ (b) 9/2 λ (c) 5/27 λ (d) 27/5 λ. Correct Answer: 1215.4Å. 3. | EduRev GATE Question is disucussed on EduRev Study Group by 133 GATE Students. The IE2 for X is? Electrons are falling to the 1-level to produce lines in the Lyman series. 3.4k VIEWS. 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. Calculate the wavelength of the lowest-energy line in the Lyman series to three significant figures. What is Lyman Series? The four other spectral line series, in addition to the Balmer series, are named after their discoverers, Theodore Lyman, A.H. Pfund, and F.S. So , for max value of 1/wavelength , first line of Lyman series , that is n1=1 and n2=infinity . 1. Lines are named sequentially starting from the longest wavelength/lowest frequency of the series, using Greek letters within each series. The first line in Lyman series has wavelength λ. are solved by group of students and teacher of JEE, which is also the largest student community of JEE. 911.2 Å. Related Questions: Ans: (a) Sol: Series Limit means Shortest possible wavelength . The wavelength of first line of lyman series i.e the electron will jump from n=1 to n=2. The Lyman series means that the final energy level is 1 which is the minimum energy level, the ground state, in other words. What is the velocity of photoelectron? The wavelength of first line of Lyman series will be 5:26 42.9k LIKES. Atoms. asked Dec 23, … As per formula , 1/wavelength = Rh ( 1/n1^2 —1/n2^2) , and E=hc/wavelength , for energy to be max , 1/wavelength must max . n 2 is the level being jumped from. 3.6k SHARES. 6.8 The first line in the Lyman series for the H atom corresponds to the n = 1 → n = 2 transition. First line is Lyman Series, where n 1 = 1, n 2 = 2. Create a New Plyalist. A stationary ion emitted a photon corresponding to a first line of the Lyman series. The Questions and Answers of The wavelength of the first line of lyman series of hydrogen is identical to that of second line of balmer series for same hydrogen like ion 'X'. Doubtnut is better on App. What is the… Correct Answer: 27/5 λ. The rest of the lines of the spectrum (all in the ultraviolet) were discovered by Lyman from 1906-1914. Further, you can put the value of Rh to get the numerical values Explanation: No explanation available. It is the transitions from higher electron orbitals to this level that release photons in the UltraViolet band of the ElectroMagnetic Spectrum. If the interaction between radiation and the electron is V = eE:r = e(Ecx + Eyy + E,z), which (n, €, m) states mix with the state (1,0,0) to give this absorption line, called Lyman a? The wavelength of the first line in Balmer series is . Class 10 Class 12. For the Balmer series, n 1 is always 2, because electrons are falling to the 2-level. Energy, ΔE=13.6( n 1 2 1 − n 2 2 1 ) eV For the first line of Lyman series: n 1 =1, n 2 =2 ΔE=13.6( 1 2 1 − 2 2 1 ) eV=10.2 eV and energy decreases as we move on to the next series. We have step-by-step solutions for your textbooks written by Bartleby experts! OR. 1:25 16.5k LIKES. Calculate the wavelength corresponding to series limit of Lyman series. 17. Add to playlist. OR. Solution for The first line of the Lyman series of the hydrogen atom emission results from a transition from the n = 2 level to the n = 1 level. The first line in each series is the transition from the next lowest number in the series to the lowest (so in the Lyman series the first line would be from n=2 to n=1) and the second line would be from from the third lowest to the lowest (in Lyman it would be n=3 to n=1) etc etc. Wave length λ = 0.8227 × 10 7 = 8.227 × 10 6 m-1 (b) Identify the region of the electromagnetic spectrum in which these lines appear. Textbook solution for Modern Physics 3rd Edition Raymond A. Serway Chapter 4 Problem 12P. The atomic number Z of hydrogen-like ion is. New questions in Chemistry. The wavelength of the first line of Lyman series for 20 times ionized sodium atom will be added 0.1 A˚ The Lyman series lies in the ultraviolet, whereas the Paschen, Brackett, and Pfund series … Explanation: No explanation available. 3.6k VIEWS. The wavelengths of the Lyman series for hydrogen are given by $$\frac{1}{\lambda}=R_{\mathrm{H}}\left(1-\frac{1}{n^{2}}\right) \qquad n=2,3,4, \ldots$$ (a) Calculate the wavelengths of the first three lines in this series. 812.2 Å . Currently only available for. 1. The first line in the spectrum of the Lyman series was discovered in 1906 by Harvard physicist Theodore Lyman, who was studying the ultraviolet spectrum of electrically excited hydrogen gas. Options (a) 1215.4Å (b) 2500Å (c) 7500Å (d) 600Å. Nov 09,2020 - If the wavelength of the first line of Lyman series of hydrogen is 1215 Å. the wavelength of the second line of the series isa)911Åb)1025Åc)1097Åd)1008ÅCorrect answer is option 'B'. And this initial energy level has to be higher than this one in order to have a transition down to it and so the first line is gonna have an initial equal to 2. 712.2 Å. (a) v 1 – v 2 = v 3 (b) v 2 – v 1 = v 3 (c) v 3 = ½ (v 1 + v 2) (d) v 2 + v 1 = v 3. 4. 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