Angular momentum of an electron in a hydrogen atom is $\frac{3h}{2\pi}$. The wavelength of this electron is approximately $...... \mathring{A}$.

In Bohr's model,the atomic radius of the first orbit is $r_0$,then the radius of the third orbit is

The wavelength for an electron in an orbit of a hydrogen atom is $10^{-9} \ m$. The principal quantum number for this electron is:

The de-Broglie wavelength of an electron in the $n^{th}$ Bohr orbit is $\lambda_n$ and the angular momentum is $J_n$,then:

The inverse square law in electrostatics is $|\vec F| = \frac{{{e^2}}}{{4\pi { \in _0}{r^2}}}$ for the force between an electron and a proton. The $\frac{1}{r^2}$ dependence of $|\vec F|$ can be understood in quantum theory as being due to the fact that the particle of light (photon) is massless. If photons had a mass $m_p$,the force would be modified to $|\vec F| = \frac{{{e^2}}}{{4\pi { \in _0}}}\left( {\frac{1}{{{r^2}}} + \frac{\lambda }{r}} \right)\left( {{e^{ - \lambda r}}} \right)$ where $\lambda = \frac{{{m_p}c}}{\hbar }$ and $\hbar = \frac{h}{{2\pi }}$. Estimate the change in the ground state energy of a $H$-atom if $m_p$ were $10^{-6}$ times the mass of an electron.

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