Question

A hydrogen atom in the ground state is excited by radiations of wavelength $975A$${}^o$. How many lines will be possible in emission spectrum?

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is C $6$

Given : $\lambda =975$ $A^o=97.5nm$

Energy of incident radiation , $E=\frac{hc}{\lambda }=\frac{1242.4}{97.5}=12.74$ eV

Energy of the state into which the electron jumps $E=-13.6-(-12.74)=-0.85$ eV

We get $-0.86=\frac{-13.6}{n^2}$ $⟹n=4$

$\therefore$ Number of emission lines $N=n(n-1)/2=4\left(3\right)/2=6$