Bohr’s second postulate implies the quantisation of:

$A$ 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 $m_p = (5/3) \times 10^{-27} \ kg$; $h/e = 4.2 \times 10^{-15} \ J \cdot s/C$; $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$; $1 \ fm = 10^{-15} \ m$)

The magnetic field at the centre due to the motion of an electron in the first Bohr orbit is $B$. The magnetic field due to the motion of an electron in the second Bohr orbit at the centre will be

$A$ hydrogen atom is in its $n^{\text{th}}$ energy state. If the de-Broglie wavelength of the electron is $\lambda$,then:

The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 10^{-11} \text{ m}$. What is the radius of the $n = 3$ orbit?

Angular momentum of an electron in a hydrogen atom is $\frac{3h}{\pi}$, then the energy of the electron is . . . . . . eV.