$A$ 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 Bohr model for the $H$-atom relies on Coulomb's law of electrostatics. Coulomb's law has not been directly verified for very short distances of the order of $\mathring{A}$. Suppose Coulomb's law between two opposite charges $+q_1$ and $-q_2$ is modified to $|\vec{F}| = \frac{q_1 q_2}{4\pi \epsilon_0} \left( \frac{1}{r^2} \right)$ for $r \ge R_0$ and $|\vec{F}| = \frac{q_1 q_2}{4\pi \epsilon_0} \left( \frac{1}{R_0^{2-\epsilon} r^{\epsilon}} \right)$ for $r < R_0$. Calculate the ground state energy of an $H$-atom,given $\epsilon = 0.1$ and $R_0 = 1 \,\mathring{A}$.

The angular momentum of the orbital electron is an integral multiple of:

If the energy of the $n^{th}$ orbit in a hydrogen atom is $E_n$,what will be the energy of the $n^{th}$ orbit in a helium ion $(He^+)$?

An electron in a hydrogen atom first jumps from the second excited state to the first excited state and then from the first excited state to the ground state. Let the ratio of wavelength,momentum,and energy of the photons in the two cases be $x, y,$ and $z$ respectively. Select the wrong answer$(s)$:

The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is ..... (radius of the first orbit of hydrogen atom $= 0.53 \ \text{Å}$). (in $\text{Å}$)