According to the classical electromagnetic theory. calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom.
Answer
We know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius \(5.3\times 10^{-11}\;m\) is \(2.2\times 10^{-6}\;m\text /s\) . Thus. the frequency of the electron moving around the proton is \(\nu=\frac v{2\mathit{\pi r}}=\dfrac{2.2\times 10^6\;ms^{-1}}{2\pi (5.3\times 10^{-11}\;m)}\) \(=6.6\times 10^{15}\mathit{Hz}\) According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal \(10\) the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is \(=6.6\times 10^{15}\mathit{Hz}\) .
A \(10\;\mathrm{km}\) satellite circles earth once every \(2\;h\) in an orbit, having a radius of \(8000\;\mathrm{km}\) . Assuming that Bohr's angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom. find the quantum number of the orbit of the satellite.
Consider the statements \(S_{1}\) and \(S_{2}\) : \(S_{1}:\) Conductivity always increases with decrease in the concentration of electrolyte. \(S_{2}:\) Molar conductivity always increases with decrease in the concentration of electrolyte. The correct option among the following is( )
Pick an appropriate option to justify the given statement: Size of an atom in Rutherford’s model is _________ the size of an atom compared to Thomson’s model. \((\)Less than, Greater than, no different than\()\)