For a hydrogen atom,when an electron jumps from $n = 2$ to $n = 1$,the wavelength of the radiation emitted is found to be $\lambda_0$. For which transition of an electron in a $He^+$ ion will the wavelength of the radiation emitted be equal to $\lambda_0$?

The orbital acceleration of an electron in a hydrogen atom is given by:

In a mixture of $H-He^{+}$ gas ($He^{+}$ is a singly ionized $He$ atom),$H$ atoms and $He^{+}$ ions are excited to their respective first excited states. Subsequently,$H$ atoms transfer their total excitation energy to $He^{+}$ ions by collisions. Assume that the Bohr model of the atom is exactly valid.
$1.$ The quantum number $n$ of the state finally populated in $He^{+}$ ions is
$(A) 2$ $(B) 3$ $(C) 4$ $(D) 5$
$2.$ The wavelength of light emitted in the visible region by $He^{+}$ ions after collisions with $H$ atoms is
$(A) 6.5 \times 10^{-7} \ m$ $(B) 5.6 \times 10^{-7} \ m$ $(C) 4.8 \times 10^{-7} \ m$ $(D) 4.0 \times 10^{-7} \ m$
$3.$ The ratio of the kinetic energy of the $n=2$ electron for the $H$ atom to that of the $He^{+}$ ion is
$(A) 1/4$ $(B) 1/2$ $(C) 1$ $(D) 2$

The ground state energy of a hydrogen atom is $-13.6 \ eV$. In this state,the potential energy is . . . . . . $eV$.

$A$ small particle of mass $m$ moves in such a way that its potential energy $U = \frac{1}{2} m \omega^2 r^2$,where $\omega$ is a constant and $r$ is the distance of the particle from the origin. Assuming Bohr's quantization of angular momentum and a circular orbit,the radius of the $n^{\text{th}}$ orbit will be proportional to:

In a hydrogen atom,which quantity is an integral multiple of $\frac{h}{2\pi}$?