Question

What is the energy of the photon that, when absorbed by a hydrogen atom, could cause the following? (a) an electronic transition from the n = 3 state to the n = 6 state

(b) an electronic transition from the n = 3 state to the n = 8 state

Answer 1

(a):
`\rm 1.133\ eV\ \ \ or\ \ \ 1.8128* 10^(-19)\ J.`

(b):
`\rm 1.298\ eV \ \ \ or \ \ \ 2.077* 10^(-19)\ J.`

Step-by-step explanation:

The energy of the photon that absorbed by a hydrogen atom causes a transition is equal to the difference in energy levels of the hydrogen atom corresponding to that transition.

According to Rydberg's formula, the energy corresponding to
`\rm n^(th)` level in hydrogen atom is given by

`\rm E_n = -(E_o)/(n^2).`

where,

`\rm E_o=13.6\ eV.`

Part (a): For the electronic transition from the n = 3 state to the n = 6 state.

The energy of the photon which cause this transition is given by

`\rm \Delta E=E_6-E_3.\\\\where,\\E_6=-(E_o)/(6^2)=-(13.6)/(36)=-0.378\ eV.\\E_3=-(E_o)/(3^2)=-(13.6)/(9)=-1.511\ eV.\\\\\therefore \Delta E = (-0.378)-(-1.511)=1.133\ eV\\or\ \ \ \Delta E = 1.133* 1.6* 10^(-19)\ J=1.8128* 10^(-19)\ J.`

Part (b): For the electronic transition from the n = 3 state to the n = 8 state.

The energy of the photon which cause this transition is given by

`\rm \Delta E=E_8-E_3.\\\\where,\\E_8=-(E_o)/(8^2)=-(13.6)/(64)=-0.2125\ eV.\\E_3=-(E_o)/(3^2)=-(13.6)/(9)=-1.511\ eV.\\\\\therefore \Delta E = (-02125)-(-1.511)=1.298\ eV\\or\ \ \ \Delta E = 1.298* 1.6* 10^(-19)\ J=2.077* 10^(-19)\ J.`