The radius of the electron's second stationary orbit in Bohr's atom is $R$. The radius of the third orbit will be:

In a hydrogen atom,if an electron in the orbit with principal quantum number $n$ jumps to the first excited state,the wavelength of the emitted photon is $\lambda$. Then the value of $n$ is (where $R$ is the Rydberg constant).

If the force between the electron in the first Bohr orbit and the nucleus (proton) in a hydrogen atom is $F$,then the force between them when the electron is in the second orbit is

Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6} \times 10^5 \text{ m s}^{-1}$. The radius of the orbit is $x \times 10^{-9} \text{ m}$. The value of $x$ is : [Take the mass of electron to be $9 \times 10^{-31} \text{ kg}$, charge of electron = $-1.6 \times 10^{-19} \text{ C}$ and $\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$]

Hydrogen $(H)$,Deuterium $(D)$,Helium $(He^+)$,and Lithium $(Li^{2+})$ emit radiation of wavelengths $\lambda_1, \lambda_2, \lambda_3,$ and $\lambda_4$ respectively during the transition from $n = 2$ to $n = 1$. Then:

As the electron in a Bohr orbit of a Hydrogen atom transitions from state $n = 2$ to $n = 1$,how do the kinetic energy $K$ and potential energy $U$ change?