Question

One of the emission spectral lines for Be31 has a wavelength of 253.4 nm for an electronic transition that begins in the state with n 5 5. What is the principal quantum number of the lower-energy state corresponding to this emission? (Hint: The Bohr model can be applied to one-electron ions. Don’t forget the Z factor: Z 5 nuclear charge 5 atomic number.)

Answer 1

Step-by-step explanation:

The given data is as follows.

`\lambda` = 253.4 nm =
`253.4 * 10^(-9)m` (as 1 nm =
`10^(-9)`)

`n_(1)` = 5,
`n_(2)` = ?

Relation between energy and wavelength is as follows.

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

=
`(6.626 * 10^(-34) Js * 3 * 10^(8) m/s)/(253.4 * 10^(-9))`

=
`0.0784 * 10^(-17)` J

=
`7.84 * 10^(-19) J`

Hence, energy released is
`7.84 * 10^(-19) J`.

Also, we known that change in energy will be as follows.

`\Delta E = -2.178 * (Z)^(2)[(1)/(n^(2)_(2)) - (1)/(n^(2)_(1))`

where, Z = atomic number of the given element

`7.84 * 10^(-19) J = -2.178 * (4)^(2)[(1)/(n^(2)_(2)) - (1)/((5)^(2))`

`(7.84 * 10^(-19) J)/(34.848) = (1)/(n^(2)_(2)) - (1)/((5)^(2))`

0.02 + 0.04 =
`(1)/(n^(2)_(1))`

`n_(1)` =
`\sqrt{(1)/(0.06)}`

= 4

Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.