$A$ light of energy $12.75 \; eV$ is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is $\frac{x}{\pi} \times 10^{-17} \; eVs$. The value of $x$ is $........$ (use $h=4.14 \times 10^{-15} \; eVs$)

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

The magnetic moment of an electron due to its orbital motion is proportional to $(n =$ principal quantum number$)$

To calculate the size of a hydrogen ion $(H^-)$ using the Bohr's model,we assume that its two electrons move in an orbit such that they are always on diametrically opposite sides of the nucleus. With each electron having the angular momentum $\hbar = h / 2\pi$ and taking electron interaction into account,the radius of the orbit in terms of the Bohr's radius of hydrogen atom $a_B = \frac{4\pi\varepsilon_0\hbar^2}{me^2}$ is

$A$ 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:

The energy of a hydrogen atom in the ground state is $-13.6 \, eV$. The energy of the $He^+$ ion in the first excited state will be .... $eV$.