Question

Applying the bohr model to a triply ionized beryllium atom (be3+,z=4), find the shortest wavelength of the lyman series for be3+.

Answer 1

`\lambda=5.69*10^(-9)m=5.69nm`

Step-by-step explanation:

The Bohr model predicts energy with values:

`h\\u = Z^2E_0((1)/(n^2)-(1)/(m^2))\\E_0=-13.6eV`

`\lambda=(hc)/(Z^2E_0)((1)/((1)/(n^2)-(1)/(m^2)))`

with z the atomic number, h the Planck,s constant, c is the speed of light and v the frequency.

The Lyman series is given by

`(1)/(\lambda)=R((1)/(1^2)-(1)/(n^2))`

R=1.097*10^7m^-1

That is, for transition between n>=2 and n=1. The shortest wavelength is obtained for n=infinity.

Hence, by replacing we have (h=4.13*10^{-15}eV , c=3*10^8 m/s ):

`\lambda=5.69*10^(-9)m=5.69nm`

hope this helps!!