Question

THE THE QUANTUM NUMBERS OF EXITED STATE OF ELECTRON IN He⁺ ION WHICH ON TRANSITION TO GROUND STATE AND FIRST EXCITED EMIT TWO PHOTONS OF WAVELENGTH 30.4 nm AND 108.5nm RESP. (R_H=1.097 x 10^-7m^-1).

Answer 1

Dear User,
Wavelength of first photon, λ₁ = 108.5 x 10^-9
Wavelength of second photon, λ₂= 30.4 x 10^-9 m​
^{$$\begin{gathered} \mathrm{Let}\mathrm{excited}\mathrm{state}\mathrm{of}{\mathrm{He}}^{+}\mathrm{be}{\mathrm{n}}_2\mathrm{to}{\mathrm{n}}_1\mathrm{and}\mathrm{then}{\mathrm{n}}_1\mathrm{to}1\mathrm{emit}\mathrm{two}\mathrm{successive}\mathrm{photon}. \\ \frac{\mathbf{1}}{{\mathbf{\lambda }}_{\mathbf{2}}}\mathbf{=}{\mathbf{R}}_{\mathbf{H}}\mathbf{.}\mathbf{}{\mathbf{z}}^{\mathbf{2}}\mathbf{[}\frac{\mathbf{1}}{{\mathbf{1}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{1}}{{{\mathbf{n}}_{\mathbf{1}}}^{\mathbf{2}}}\mathbf{]} \\ \frac{1}{30.4\times {10}^{-9}}=1.097\times {10}^{-7}\left(2{)}^2\right[\frac{1}{1}-\frac{1}{{{\mathrm{n}}_1}^2}] \\ \mathrm{Therefore},{\mathrm{n}}_1=2 \\ \mathrm{Now},\mathrm{for}{\mathrm{\lambda }}_1{\mathrm{n}}_1=2 \\ \frac{1}{{\mathrm{\lambda }}_1}={\mathrm{R}}_{\mathrm{H}}.{\mathrm{z}}^2[\frac{1}{2^2}-\frac{1}{{{\mathrm{n}}_2}^2}] \\ \frac{1}{108.5\times {10}^{-9}}=1.097\times {10}^{-7}(2{)}^2[\frac{1}{4}-\frac{1}{{{\mathrm{n}}_2}^2}] \\ \mathrm{Therefore},{\mathrm{n}}_2=5 \\ \mathrm{Hence},\mathrm{excited}\mathrm{state}\mathrm{of}{\mathrm{He}}^{+}\mathrm{is}5\mathrm{th}\mathrm{orbit}\mathrm{or}\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}\mathbf{the}\mathbf{}\mathbf{quantum}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{excited}\mathbf{}\mathbf{state}\mathbf{}\mathbf{is}\mathbf{}\mathbf{5}\mathbf{.} \end{gathered}$$}m