In a hypothetical atom, mass of electron is doubled, value
of atomic number is `Z = 4`. Find wavelength of photon when this electron jumps
from 3rd excited state to 2nd orbit.

Answer

A

Explanation

`E prop (Z^2)/(n^2)`prop m`
Mass is doubled, `Z = 4` and 3rd excited state means `n =4`, second orbit means `n = 2` for these`
values, we have. `E_4 =- 13.6xx2((4)/(4))^2 =- 27.2eV` and `E_2 =-13.6xx2((4)/(2))^2 =- 108.8eV`
`lmabda(inÅ)= (12375)/(DeltaE(in eV)) = (12375)/(E_4 - E_2)` `=(12375)/(-27.2+108.8)`
`:. lambda = 151.65Å`