1.If the shortest wavelength of the Lyman series of a hydrogen atom is xxx, the wavelength of the first member of the Balmer series isa.9x25\dfrac{9x}{25}259xb.36x5\dfrac{36x}{5}536xc.5x9\dfrac{5x}{9}95xd.5x36\dfrac{5x}{36}365xLogin to continueOnly logged in users canattempt or see the solution.
