Minimum excitation potential of Bohr's first orbit in hydrogen atom is.....$V$

$A$ diatomic molecule is made of two masses $m_1$ and $m_2$ separated by a distance $r$. Applying the Bohr's quantization rule for angular momentum,calculate its rotational kinetic energy. It is given by the formula:

Obtain the equations for the frequency of radiation and the wave number when an electron makes a transition from a higher energy state to a lower energy state in a hydrogen atom.

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 $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$)

Energy levels $A, B, C$ of a certain atom correspond to increasing values of energy,i.e.,$E_A < E_B < E_C$. If $\lambda_1, \lambda_2, \lambda_3$ are the wavelengths of radiations corresponding to the transitions $C$ to $B$,$B$ to $A$,and $C$ to $A$ respectively,which of the following statements is correct?

The de-Broglie wavelength $(\lambda_B)$ associated with the electron orbiting in the second excited state of a hydrogen atom is related to that in the ground state $(\lambda_G)$ by: