What is the energy difference between the two energy levels involved in the emission that results in this spectral line? H-alpha light is the brightest hydrogen line in the visible spectral range. We get Balmer series of the â¦ The first line of Balmer series has wavelength 6563 Å. The wavelength of the first member of Balmer series in the hydrogen spectrum is 6563A o.Calculate the wavelength of the first member of lyman series in the same spectrum. For first member of Balmer series wave length is â¦ View All Answers (1) S safeer. The wavelength of first line of balmer series in hydrogen spectrum is 6563 A. Using Rydberg formula, calculate the wavelengths of the spectral lines of the first member of the Lyman series and of the Balmer series. Then the wavelength of the second member is. What will be the wavelength of the first member of Lyman series. Then the wavelength (in Å) of the first member of Lyman series in the - 11005293 The wavelength of the first member of the Balmer series in hydrogen spectrum is x Å. What will be the wavelength of the first member of Lyman series (a) 1215.4 Å (b) 2500 Å (c) 7500 Å (d) 600 Å Calculate the value of Rydberg constant if the wavelength of the first member of Balmer series in the hydrogen spectrum is 6563 amstrong. If the wavelength of first member of Balmer series of hydrogen spectrum is 6564 A°, the wavelength of second member of Balmer series will be: (A) 121 AOC fires back at critics of her Vanity Fair photo shoot mâ»¹ A/C to question, here it is given that first member of balmer series of hydrogen atom has wavelength of 656.3 â¦ H-alpha (HÎ±) is a specific deep-red visible spectral line in the Balmer series with a wavelength of 656.28 nm in air; it occurs when a hydrogen electron falls from its third to second lowest energy level. Find the wavelength of first line of lyman series in the same spectrum. the shortest line of Lyman series p = 1 and n = â Balmer Series: If the transition of electron takes place from any higher orbit (principal quantum number = 3, 4, 5, â¦) to the second orbit (principal quantum number = 2). The first line of the Balmer series occurs at a wavelength of 656.3 nm. Also find the wavelength of the first member of Lyman series in the same spectrum. A. The first line of Balmer series has wavelength 6563 â«. Solution Show Solution The Rydberg formula for the spectrum of the hydrogen atom is given below: