Question

The series in the He+He+ spectrum that corresponds to the set of transitions where the electron falls from a higher level to the ????f=4nf=4 state is called the Pickering series, an important series in solar astronomy. Calculate the Pickering series wavelength associated with the excited state ????i=8ni=8 .

Answer 1

Step-by-step explanation:

According to the Rydberg formula,

`(1)/(\lambda) = -R_(H)Z^(2) [(1)/(n^(2)_(i)) - (1)/(n^(2)_(f))]`

where,
`\lambda` = wavelength

`R_(H)` = Rydberg constant =
`1.097 * 10^(7) m^(-1)`

Z = atomic number (for He atomic number is 2)

`n_(f)` = 4,
`n_(i)` = 8

Hence, putting the given values into the above formula as follows.

`(1)/(\lambda) = -R_(H)Z^(2) [(1)/(n^(2)_(i)) - (1)/(n^(2)_(f))]`

=
`-1.097 * 10^(7) * (2)^(2) [(1)/((8)^(2)) - (1)/((4)^(2))]`

=
`0.2056 * 10^(7) m^(-1)`

`\lambda = (1)/(0.2056 * 10^(7) m^(-1))`

=
`4.86 * 10^(-7) m`

= 486 nm (as
`1 * 10^(-9)` = 1 nm)

Thus, we can conclude that the Pickering series wavelength associated with the excited state is 486 nm.