Find the kinetic energy, potential energy and total energy in the first and second orbit of the hydrogen atom, if potential energy in the first orbit is taken to be zero.

Now U1 = 0, i.e., potential energy has been increased by 27.20 eV. So, we will increase U and E in all energy states by 27.20 eV while kinetic energy will remain unchanged. Changed values in tabular form are as under.
$\begin{array}{l}{\mathrm{E}}_1=-13.60\mathrm{eV}\\ {\mathrm{K}}_1=-{\mathrm{E}}_1=13.60\mathrm{eV}\\ {\mathrm{U}}_1=2{\mathrm{E}}_1=-27.20\mathrm{eV}\\ {\mathrm{E}}_2=\frac{{\mathrm{E}}_1}{(2{)}^2}=-3.40\mathrm{eV}\\ {\mathrm{K}}_2=3.40\mathrm{eV}\end{array}$
and ${\mathrm{U}}_2=-6.80\mathrm{eV}$