State the dimension of the nucleus from Rutherford experiment.
An alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze.$ Then the distance of closest approach for the alpha nucleus will be proportional to
If in nature there may not be an element for which the principal quantum number $n > 4,$ then the total possible number of elements will be
In third orbit of hydrogen atom, de Broglie wavelength of electron is $\lambda $ then radius of third orbit is
An $\alpha$- particle of $5\ MeV$ energy strikes with a nucleus of uranium at stationary at an scattering angle of $180^o$. The nearest distance upto which $\alpha$- particle reaches the nucleus will be of the order of
A proton is fired from very far away towards a nucleus with charge $Q=120 \ e$, where $e$ is the electronic charge. It makes a closest approach of $10 \ fm$ to the nucleus. The de Brogle wavelength (in units of $fm$ ) of the proton at its start is :
(take the proton mass, $m _0=(5 / 3) \times 10^{-27} kg , h / e =4.2 \times 10^{-15} J / s / C ; \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 m / F ; 1 fm =10^{-15} m$ )