If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of de-Broglie waves for electrons in the first and second Bohr orbits in a hydrogen atom,then the ratio $\left(\frac{\lambda_{1}}{\lambda_{2}}\right)$ is equal to:

$A$ photon of wavelength $300 \ nm$ interacts with a stationary hydrogen atom in the ground state. During the interaction, the whole energy of the photon is transferred to the electron of the atom. State which possibility is correct. (Consider, Planck constant $= 4 \times 10^{-15} \ eV \cdot s$, velocity of light $= 3 \times 10^{8} \ m/s$, ionisation energy of hydrogen $= 13.6 \ eV$)

$A$ $\mu$-meson of charge '$e$',mass $208 m_e$ moves in a circular orbit around a heavy nucleus having charge $+3e$. The quantum state '$n$' for which the radius of the orbit is the same as that of the first Bohr orbit for a hydrogen atom is [approximately]:

When a hydrogen atom is excited by radiation of wavelength $975 \, \mathring{A}$,the maximum and minimum wavelengths of the emitted radiation are respectively:

The gravitational force between a $H$-atom and another particle of mass $m$ is given by Newton's law $F = G\frac{M m}{r^2}$. Here,$M$ represents:

The electron of a hydrogen atom makes a transition from the $(n + 1)^{th}$ orbit to the $n^{th}$ orbit. For large $n$,the wavelength of the emitted radiation is proportional to: