Question

Like He, the Li2+ ion is a single-electron system (Problem 5.94). What wavelength of light in nm must be absorbed to

promote the electron in Li2+ from n = 1 to n = 4?​

Answer 1

Answer:
`1.08* 10^(-8)m`

Step-by-step explanation:

Let
`E_(n)` be the energy of the electron in
`n`th orbit.

According to Bohr's model,

`E_(n)=(-kz^(2))/(n^(2))`

where
`k=2.179* 10^(-18)J`

`Z` is the atomic number

`n` is the orbit number.

Given,
`z=3`

Energy required for transition from
`n=1` to
`n=4` is
`(k(3)^(2))/(1^(2))-(k(3)^(2))/(4^(2))=(15k* 9)/(16) =18.38* 10^(-18)J`

Since,wave length is
`(hc)/(E)`

where
`h` is the plancks constant.

`c` is the speed of light.

`c=3* 10^(8)\\h=6.63* 10^(-34)m^(2)KgS^(-1)`

So,wave length is
`(6.63* 10^(-34) * 3* 10^(8))/(18.38* 10^(-18)) =1.08* 10^(-8)m`