JEE Main 202431 Jan, Morning shiftPhysicsEasyAtoms and Nuclei183× since 2002If the wavelength of the first member of Lyman series of hydrogen is λ\lambdaλ. The wavelength of the second member will beAOption A. 275λ\frac{27}{5} \lambda527λBOption B. 527λ\frac{5}{27} \lambda275λCOption C. 2732λ\frac{27}{32} \lambda3227λDOption D. 3227λ\frac{32}{27} \lambda2732λCheckSkip · show solution
