What is the lowest energy of the spectral line emitted in the Lyman series of hydrogen atom?

We know that $\Delta E=hc{R}_H\times Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
For the lowest energy of spectral line in the lyman series.
$n_1=1,n_2=2$ 
Hence, $\Delta E=hc{R}_H\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}\right)=hc{R}_H\left(\frac{1}{1}-\frac{1}{4}\right)=\frac{3hc{R}_H}{4}$