Question

The energy levels of hydrogenlike one-electron ions of atomic number Z differ from those of hydrogen by a factor of Z^2. Predict the wavelength of the 2s--->1s transition in He+.

Answer 1

`\mathbf{\lambda \simeq 3.039 * 10^(-8) \ m}`

Step-by-step explanation:

For a hydrogen-like atom, the spectral line wavelength can be computed by using the formula:

`\bar v = Z^2 R_H \Big((1)/(n_f^2)-(1)/(n_i^2)\Big)`

where:

emitted radiation of the wavenumber
`\bar v` = ???

atomic no of helium Z = 2

Rydberg's constant
`R_H = 1.097*10^7 \ m^(-1)`

the initial energy of the principal quantum
`n_1` = 2

the initial energy of the principal quantum
`n_1` = 2

Now, the emitted radiation of the wavenumber can be computed as:

`\bar v = (2)^2 (1.097*10^7 \ m^(-1) ) \Big((1)/(1^2)-(1)/(2^2)\Big)`

`\bar v = 3.291 * 10^ 7/m`

Now, the wavelength for the transition can be computed by using the relation between the wavelength λ and the emitted radiation of the wavenumber
`\bar v`, which is:

`\bar v = (1)/(\lambda)`

`\lambda = (1)/(\bar v)`

`\lambda = (1)/(3.291 * 10^(7))* (m)/(1)`

`\mathbf{\lambda =3.03859 * 10^(-8) \ m}`

`\mathbf{\lambda \simeq 3.039 * 10^(-8) \ m}`