According to Bohr's theory,the time-averaged magnetic field at the centre (i.e.,nucleus) of a hydrogen atom due to the motion of electrons in the $n^{th}$ orbit is proportional to ($n =$ principal quantum number).

If an electron is revolving in its Bohr orbit having Bohr radius of $0.529 Å$,then the radius of the third orbit is

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 $R_0 = 1 \,\mathring{A}$.

An electron has a mass of $9.1 \times 10^{-31} \ kg$. It revolves around the nucleus in a circular orbit of radius $0.529 \times 10^{-10} \ m$ at a speed of $2.2 \times 10^6 \ m/s$. The magnitude of its linear momentum in this motion is:

Speed of an electron in Bohr's $7^{\text{th}}$ orbit for Hydrogen atom is $3.6 \times 10^6\,m/s$. The corresponding speed of the electron in $3^{\text{rd}}$ orbit,in $m/s$ is $........\times 10^6$.

$A$ hydrogen atom in its ground state absorbs $10.2 \ eV$ of energy. The orbital angular momentum is increased by (Given Planck constant $h = 6.6 \times 10^{-34} \ J \cdot s$)