$A$ particle of mass $200 \, MeV/c^2$ collides with a hydrogen atom at rest. Soon after the collision,the particle comes to rest,and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in $eV$) is $\frac{N}{4}$. The value of $N$ is: (Given the mass of the hydrogen atom to be $1 \, GeV/c^2$)

The de-Broglie wavelength of the electron in the first Bohr orbit of the hydrogen atom is

If the de-Broglie wavelength of an electron in the first Bohr orbit is $\lambda$,then the minimum radial distance between the electrons in the first and second Bohr orbits is:

$A$ proton is fired from a very large distance towards a nucleus with charge $Q = 120e$,where $e$ is the elementary charge. It reaches a distance of closest approach of $10 \ fm$. The de Broglie wavelength of the proton at its initial position is (in $fm$). (Given: mass of proton $m_p = (5/3) \times 10^{-27} \ kg$; $h/e = 4.2 \times 10^{-15} \ J \cdot s/C$; $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$; $1 \ fm = 10^{-15} \ m$)

An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transitions. An electron is moving in an orbit of another hydrogen atom from which there can be a maximum of three transitions. The ratio of the velocities of the electron in these two orbits is

The radius of a hydrogen atom in its ground state is $5.3 \times 10^{-11} \ m$. After collision with an electron,it is found to have a radius of $21.2 \times 10^{-11} \ m$. What is the principal quantum number $n$ of the final state of the atom?