The total energy of an electron in the $n^{th}$ orbit of a hydrogen atom is $E_n$. The total energy of an electron in the $n^{th}$ orbit of a $He^+$ ion is .........

$A$ doubly charged lithium atom $(Li^{++})$ is a hydrogen-like species with atomic number $Z = 3$. The wavelength of the radiation required to excite an electron from the first to the third Bohr orbit in $Li^{++}$ is.......$\mathring{A}$ (Ionization energy of hydrogen atom is $13.6\,eV$).

The de Broglie wavelength of the electron in the ground state is $\lambda_1$ and that in the $n = 3$ level is $\lambda_3$. Then $\lambda_3$ is given by:

The kinetic energy of an electron revolving around a nucleus will be

$A$ 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:

The ratio of the areas of the electron orbits for the second excited state to the first excited state for the hydrogen atom is