Question

The shortest wavelength of Lyman series of Hydrogen atom is $\lambda$. If an electron jumps from a higher energy level to ground state and emits a photon of wavelength 9$\lambda$/8, the principal quantum number of that higher energy level is

• A. 2
• B. 3
• C. 4
• D. 9

Answer 1

The correct option is B

3

$\frac{1}{\lambda }=R(\frac{1}{n^2}-\frac{1}{m^2}).Forshortestwavelengthm=\infty .ThisgivesR=\frac{1}{\lambda }.\frac{8}{9\lambda }=\frac{1}{\lambda }(\frac{1}{1^2}-\frac{1}{m^2}).Solvethisyougetm=3\}$