Solve 10.

Dear student

Energy in a state= KE+PE
KE is positive
PE for electron is negative

negative sign thus signifies that PE of electron in hat state is more than KE in magnitude.

If total energy in the state is -13.6eV;
Let the difference of energy in first and second state be dE (eV)

​Then amount of energy required to excite the electron to second state = -13,6+dE (eV)

dE can be found out by using raydberg' s formula

$$\begin{gathered} E=hc{R}_H(\frac{1}{1^2}-\frac{1}{2^2}) \\ h=planks cons\tan t \\ c=speed of light \\ {R}_H=rydberg cons\tan t \end{gathered}$$
Regards