The ground state energy of a hydrogen atom is $-13.6 \text{ eV}$. What is the ratio of the kinetic energy to the potential energy of the electron in this state?

The acceleration of an electron in the first orbit of the hydrogen atom $(Z = 1)$ is

The ratio of the speed of the electron in the first Bohr orbit of hydrogen and the speed of light is equal to (where $e, h$ and $c$ have their usual meanings).

Match the following List-$I$ with List-$II$ in connection with Bohr's atomic model.

$A$. Speed of revolution of electron	$i$. $\frac{1}{4 \pi \varepsilon_0} \frac{2 \pi Z e^2}{n h}$
$B$. Kinetic energy	$ii$. $-\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$C$. Total energy	$iii$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$D$. Frequency	$iv$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}$

If one takes into account the finite mass of the proton,then the correction to the binding energy of the hydrogen atom is approximately (take,mass of proton $= 1.60 \times 10^{-27} \, kg$ and mass of electron $= 9.10 \times 10^{-31} \, kg$) (in $\%$)

What is ground state?