In the Bohr model of a hydrogen atom,the electrostatic force on the electron depends on the principal quantum number $n$ as:

$A$ sample of hydrogen-like atoms produces an emission spectrum consisting of $10$ wavelengths arising from all possible transitions. During this process,the maximum angular momentum change for an electron transitioning from a higher energy level to a lower energy level is:

If the radius of the first Bohr orbit is $r$,then the de-Broglie wavelength of the electron in the $4^{\text{th}}$ orbit will be:

Which state of triply ionised beryllium $(Be^{3+})$ has the same orbital radius as that of the ground state of hydrogen?

In a hydrogen atom in its ground state,the first Bohr orbit has radius $r_1$. When the atom is raised to one of its excited states,the electron's orbital velocity becomes one-third of its initial value. The radius of that orbit is: (in $r_1$)

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 ........... $\%$