The electron of a hydrogen atom makes a transition from the $(n + 1)^{th}$ orbit to the $n^{th}$ orbit. For large $n$,the wavelength of the emitted radiation is proportional to:

In a hydrogen atom,the electron revolves around the nucleus in an orbit of radius $0.53 \times 10^{-10} \, m$. The electrical potential produced by the nucleus at the position of the electron is......$V$

Assertion : Bohr had to postulate that the electrons in stationary orbits around the nucleus do not radiate.
Reason : According to classical physics all moving electrons radiate.

Using the formula for the radius of the $n^{th}$ orbit $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$,derive an expression for the total energy of an electron in the $n^{th}$ Bohr orbit.

When an electron jumps from the orbit $n=2$ to $n=4$,the wavelength of the radiation absorbed will be ($R$ is Rydberg's constant).

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).