Question

Q 15/30 An electron of He* ion make a transition
from 1st excited state to 3rd excited state. Calculate
the energy of photon absorbed in this process.
10.2 eV
3.4 eV
6.8 eV
1.51 eV​

Answer 1

The energy of photon absorbed in this process is 10.2 eV.

Explanation:

• Excited state means energy level of the atom, ion or molecule.
• At ground state the energy level is low.
• After the atom was observed energy, the atom may be jumped from ground state to higher energy level.
• So. it is called excited state.
• Excited site is the non stable site, whenever the atom emited the energy, the atoms comes into higher state to lower state.

To Learn More...

1.Radius of fourth excited state of He+ ion is

(1) 661.25 pm

(2) 6.6125 pm

(3) 423.2 pm

(4) 896.5 pm​

https://brainly.in/question/11129473

Answer 2

$\huge \tt 10.2 \: ev$