In the Bohr model of the hydrogen atom,the linear momentum of the electron depends on the principal quantum number $n$ as:

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

For an electron moving in the $n^{\text{th}}$ Bohr orbit,the de Broglie wavelength of the electron 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?

$A$ $\mu$-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]:

The energy required to remove the electron from a singly ionized Helium atom is $2.2$ times the energy required to remove an electron from a Helium atom. The total energy required to ionize the Helium atom completely is......$eV$