(d) The wavelength of the first of Lyman series for hydrogen is identical to that of the second line of Balmer series for some hydrogen-like ion X. Okay. The 2nd 1 would be from any coastal tree to any close to what? The wavelength of the second spectral line in the Balmer series of singly - ionized helium atom is: And B we have the end equals three, and we have equals one. 46, Page 280 (i) Wavelength of second member of Lyman series : n 1 = 1, n 2 = 3 ∴ It lies in ultra violet region. And this is gonna be the HC is actually people 1.2 four times 10 to the third, uh, e v over una meters, and you should be able to get that constant here because it's a constant. Then we have e two, which is negative 13.6 times two squared be. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A ˚. Textbook solution for University Physics Volume 3 17th Edition William Moebs Chapter 6 Problem 92P. Click hereto get an answer to your question ️ Taking the wavelength of first Balmer line in hydrogen spectrum ( n = 3 to n = 2 ) as 660nm , the wavelength of the 2nd Balmer line ( n = 4 to n = 2 ) will be: (1) $(a)$ Determine the wavelength of the second Balmer line $(n=4 \text { to } n=2 \text { transition) using Fig. } Using this, we can find he we think after sc meter fold one right. NCERT NCERT Exemplar NCERT Fingertips Errorless Vol … Since we're dealing with TV, we should get for it. There we go. So this energy for the, uh, excited state and it goes to tree minus and it goes to one should get 102 centimeters. Different lines of Balmer series area l . line indicates transition from 4 --> 2. line indicates transition from 3 -->2. Basically, we're looking at any transition from and into any close to two any m prime to any question to So the second. And the first proper part eight. 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