The ionisation potential of $H$-atom is $13.6 \, eV$. When it is excited from the ground state by monochromatic radiation of $970.6 \, \mathring{A}$,the number of emission lines will be (according to Bohr's theory):

Assume that there is no repulsive force between the electrons in an atom,but the force between positive and negative charges is given by Coulomb's law as usual. Under such circumstances,calculate the ground state energy of a $He$ atom.

In a hydrogen atom,the electron is moving around the nucleus with a velocity of $2.18 \times 10^{6} \; m/s$ in an orbit of radius $0.528 \; \mathring{A}$. The acceleration of the electron is:

When an electron jumps from the orbit $n=2$ to $n=4$,the wavelength of the radiation absorbed will be ($R$ is Rydberg's constant).

$A$ 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 radius of a certain orbit of a hydrogen atom is $8.48 \mathring{A}$. If the energy of the electron in this orbit is $E/x$,then $x = . . . .$
(Given $a_0 = 0.529 \mathring{A}$,$E =$ energy of the electron in the ground state)