An electron of $H e^{+}$ ion make a transition from $1^{s t}$ excited state to $3^{r d}$ excited state.Calculate the energy of photon absorbed in this process.

Open in App

Solution

$\begin{matrix} E = \frac{- 13.6 Z^{2}}{h^{2}} E_{1} = \frac{- 13.6 \times {(2)}^{2}}{{(1)}^{2}} = - 544 e v E_{3} = \frac{- 13.6 \times 2^{2}}{{(3)}^{2}} = \frac{- 54.4}{9} e v e n e r g y a b s o r v e d = 54.4 (\frac{π 8}{9}) 48.35 e v . \end{matrix}$

Suggest Corrections

Similar questions

Q. In a sample of hydrogen atom in ground state electrons make transition from ground state to a particular excited state where path length is five times de-broglie wavelength, electrons make back transition to the ground state producing all possible photons. If photon having 2nd highest energy of this sample used to excite the electron in a particular excited state of

L i^{2 +}

ion then find the final excited state of

L i^{2 +}

ion

During the emission spectrum of electron in $H e^{+}$ ion, the electron on transition from same higher energy state drops to the ground state and first excited state emits two photon of $λ_{1}$ =304 $A^{\circ}$ and $λ_{2}$ =1085 $A^{\circ}$ respectively.The total number of photons emitted during the transition of electrons from highest excited to all lower energy level(Possible)