In third orbit of hydrogen atom, de Broglie wavelength of electron is $\lambda $ then radius of third orbit is
$3\lambda $
$\lambda $
$\frac{{3\lambda }}{{2 }}$
$\frac{{\lambda }}{{2 }}$
The following diagram indicates the energy levels of a certain atom when the system moves from $4E$ level to $E$. A photon of wavelength $\lambda _1$ is emitted. The wavelength of photon produced during it's transition from $\frac{7}{3}E$ level to $E$ is $\lambda_2$. The ratio $\frac{{{\lambda _1}}}{{{\lambda _2}}}$ will be
Assertion $(A)$ : The magnetic moment $(\mu)$ of an electron revolving around the nucleus decreases with increasing principle quantum number $(n)$.
Reason $(R)$ : Magnetic moment of the revolving electron, $\mu \propto n$.
Energy levels $A, B, C$ of a certain atom correspond to increasing values of energy i.e., $E_A < E_B < E_C$. If ${\lambda _1},{\lambda _2},{\lambda _3}$ are the wavelengths of radiation corresponding to the transition $C$ to $B, B$ to $A$ and $C$ to $A$ respectively, which of the following relation is correct ?
The radius of a nucleus is given by $r_0 A^{1 / 3}$, where $r_0=1.3 \times 10^{-15} \,m$ and $A$ is the mass number of the nucleus. The lead nucleus has $A=206$. The electrostatic force between two protons in this nucleus is approximately ................ $N$