Question

Bohr Model: an electron in a doubly ionized lithium atom +2Li(three protons in the nucleus) makes a transition from the 풏=ퟏto the 풏=ퟑlevel with an associated photon. a)Determine the photon energy associated with this transition. (10pts)

Answer 1

`\lambda =113.74{\AA}`

Step-by-step explanation:

Electron is jumping from n=1 to n=3

and for Li Z=3;

`E_n=-13.6* (Z^2)/(n^2)`

`\Delta E=-13.6* Z^2[(1)/(n_3^2) -(1)/(n_1^2)]`

`n_3=3 \,and \,n_1=1`

after solving we get;

`\Delta E=-13.6* 8 \,ev`

for the photon energy is calculated by the following equation

`\Delta E=(hc)/(\lambda)`

for easy calculation;

`\Delta E=(12375)/(\lambda) {\AA}`

where energy is taken in electron volt

after puting value of
`\Delta E` we get,

`\lambda =113.74{\AA}`