Answer:
[tex]\lambda=5.69*10^{-9}m=5.69nm[/tex]
Explanation:
The Bohr model predicts energy with values:
[tex]h\nu = Z^2E_0(\frac{1}{n^2}-\frac{1}{m^2})\\E_0=-13.6eV[/tex]
[tex]\lambda=\frac{hc}{Z^2E_0}(\frac{1}{\frac{1}{n^2}-\frac{1}{m^2}})[/tex]
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
[tex]\frac{1}{\lambda}=R(\frac{1}{1^2}-\frac{1}{n^2})[/tex]
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 ):
[tex]\lambda=5.69*10^{-9}m=5.69nm[/tex]
hope this helps!!
