An electron in the hydrogen atom jumps from $n^{th}$ energy state to the ground state. The wavelength so emitted illuminates a photosensitive material having work function $2.65 \text{ eV}$. If the maximum kinetic energy of the emitted photoelectrons is $10.1 \text{ eV}$, then the value of '$n$' is

If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of de-Broglie waves for electrons in the first and second Bohr orbits in a hydrogen atom,then the ratio $\left(\frac{\lambda_{1}}{\lambda_{2}}\right)$ is equal to:

The radius of the orbit of an electron in a Hydrogen-like atom is $4.5 a_0$,where $a_0$ is the Bohr radius. Its orbital angular momentum is $\frac{3h}{2\pi}$. It is given that $h$ is Planck constant and $R$ is Rydberg constant. The possible wavelength$(s)$,when the atom de-excites,is (are) :
$(A)$ $\frac{9}{32R}$ $(B)$ $\frac{9}{16R}$ $(C)$ $\frac{9}{5R}$ $(D)$ $\frac{4}{3R}$

If $m$ is the mass of an electron,$v$ is its velocity,and $r$ is the radius of a stationary circular orbit around a nucleus with charge $Ze$,then from Bohr's postulate,the kinetic energy $K = \frac{1}{2}mv^2$ of the electron in the $C.G.S.$ system is equal to:

An electron in a hydrogen atom,after absorbing energy photons,can jump between energy states $n_1$ and $n_2$ $(n_2 > n_1)$. It then returns to the ground state,emitting six different wavelengths in the emission spectrum. The energy of the emitted photons can be equal to,less than,or greater than the absorbed photon energy. Determine $n_1$ and $n_2$.

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