The radius of the first orbit in an $H$-atom is '$a_0$'. Then,the de-Broglie wavelength of the electron in the third orbit is: (in $\pi a_0$)

The product of angular speed and tangential speed of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is:

The kinetic energy of the electron in an orbit of radius $r$ in a hydrogen atom is ($e =$ electronic charge).

In the first excited state of a hydrogen atom, the energy of its electron is $-3.4 \text{ eV}$. The radial distance of the electron from the hydrogen nucleus in this case is approximately: (Take $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$, $e = 1.6 \times 10^{-19} \text{ C}$ and $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$)

Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6} \times 10^5 \text{ m s}^{-1}$. The radius of the orbit is $x \times 10^{-9} \text{ m}$. The value of $x$ is : [Take the mass of electron to be $9 \times 10^{-31} \text{ kg}$, charge of electron = $-1.6 \times 10^{-19} \text{ C}$ and $\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$]

In a hypothetical Bohr hydrogen atom, if the mass of the electron is doubled, then the energy of the electron in the first orbit is: (in $\text{ eV}$)