If the energy of a hydrogen atom in the $n^{th}$ orbit is ${E_n}$,then the energy in the $n^{th}$ orbit of a singly ionized helium atom will be:

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 ratio of the radius of the first Bohr orbit to that of the second Bohr orbit of the orbital electron is

What is the angular momentum of an electron in Bohr's hydrogen atom whose energy is $-3.4 \,eV$?

Experimentally it is found that $12.8 \, eV$ energy is required to separate a hydrogen atom into a proton and an electron. The orbital radius of the electron in this hydrogen atom is $\frac{9}{x} \times 10^{-10} \, m$. The value of $x$ is (Given: $1 \, eV = 1.6 \times 10^{-19} \, J$,$\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, Nm^2/C^2$,and electronic charge $e = 1.6 \times 10^{-19} \, C$)

An electron has a mass of $9.1 \times 10^{-31} \ kg$. It revolves around the nucleus in a circular orbit of radius $0.529 \times 10^{-10} \ m$ at a speed of $2.2 \times 10^6 \ m/s$. The magnitude of its linear momentum in this motion is: