If the electron in a hydrogen atom jumps from an orbit with level $n_1=2$ to an orbit with level $n_2=1$, the emitted radiation has a wavelength given by

In a hydrogen atom, the electron makes $6.6 \times 10^{15}$ revolutions per second around the nucleus in an orbit of radius $0.5 \times 10^{-10} \, m$. This is equivalent to a current of nearly:

In a hydrogen-like ion,the energy difference between the $2^{\text{nd}}$ excitation state and the ground state is $108.8 \ eV$. The atomic number of the ion is:

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

The Bohr model for the $H$-atom relies on Coulomb's law of electrostatics. Coulomb's law has not been directly verified for very short distances of the order of $\mathring{A}$. Suppose Coulomb's law between two opposite charges $+q_1$ and $-q_2$ is modified to $|\vec{F}| = \frac{q_1 q_2}{4\pi \epsilon_0} \left( \frac{1}{r^2} \right)$ for $r \ge R_0$ and $|\vec{F}| = \frac{q_1 q_2}{4\pi \epsilon_0} \left( \frac{1}{R_0^{2-\epsilon} r^{\epsilon}} \right)$ for $r < R_0$. Calculate the ground state energy of an $H$-atom,given $\epsilon = 0.1$ and $R_0 = 1 \,\mathring{A}$.

In Bohr's atomic model,the electron is assumed to revolve in a circular orbit of radius $0.5 \times 10^{-10} \; m$. If the speed of the electron is $2.2 \times 10^{6} \; m/s$,then the current associated with the electron will be $.... \times 10^{-2} \; mA$. [Take $\pi$ as $\frac{22}{7}$]