Hydrogen-like ion has the wavelength difference between the first lines of Balmer and lyman series equal to 33.4nm. The atomic number of that ion equal to RH=1.1×107m−1

Given,λB−λL=334 nmAtomic number, Z = ?Use the formula,1λ=RH1n12−1n22×Z2Where,  l wave length of radiationRH = Rydberg's constantn2 = Higher energy level = 3n1 = lower energy level = 2⇒1λ1=R122-132Z2⇒λ1=365 RZ21λ2=R112-122∴λ2=43 RZ2∴Δλ=λ1-λ2=365RHZ2-43RHZ2⇒334=365-43×1RH×1Z2⇒Z2=5.86×912334=16∴Z=4