In a hydrogen atom,two electrons are in orbits of radii $r_0$ and $4r_0$. What is the ratio of their frequencies of revolution around the nucleus?

In a hydrogen atom,if an electron in the orbit with principal quantum number $n$ jumps to the first excited state,the wavelength of the emitted photon is $\lambda$. Then the value of $n$ is (where $R$ is the Rydberg constant).

An electron in an excited state of $Li^{2+}$ ion has angular momentum $\frac{3 h}{2 \pi}$. The de-Broglie wavelength of the electron in this state is $p \pi a_{0}$ (where, $a_{0} = \text{Bohr radius}$). The value of $p$ is

The first four spectral lines in the Lyman series of a $H$ atom are $\lambda = 1218 \, \mathring{A}, 1028 \, \mathring{A}, 974.3 \, \mathring{A}, 951.4 \, \mathring{A}$. If instead of Hydrogen,we consider Deuterium,calculate the shift in the wavelength of these lines.

The orbital acceleration of an electron in a hydrogen atom is given by:

The emission series of hydrogen atom is given by $\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$ where,$R$ is the Rydberg constant. For a transition from $n_{2}$ to $n_{1}$,the relative change $\Delta \lambda / \lambda$ in the emission wavelength,if hydrogen is replaced by deuterium (assume that,the mass of proton and neutron are the same and approximately $2000$ times larger than that of electrons) is ........... $\%$