Question
When the electron orbiting in hydrogen atom in its ground state moves to the third excited state, show how the de Broglie wavelength associated with it would be affected.
Alternate Answer
As $2\pi\text{r}_{n} = \text{n}\lambda ; \lambda = \frac{2\pi\text{r}_{n}}{\text{n}}(\lambda\propto\frac{\text{r}_{n}}{\text{n}})$ $\text{r}_{n}\propto\text{n}^{2}$ $\therefore \lambda\propto\frac{\text{n}^{2}}{\text{n}}\Rightarrow\lambda\propto\text{n}$ $\therefore$ 𝑑𝑒 Broglie wavelength will increase.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Mass of nucleus = mA
$\text{Volume of nucleus }=\frac{4}{3}\pi\text{R}^3=\frac{4}{3}\pi\big(\text{R}_0\text{A}\frac{1}{3}\big)=\frac{4}{3}\pi\text{R}^3_0\text{A}$
$\therefore\text{Nuclear density},\rho\text{nu}=\frac{\text{Mass of nucleus}}{\text{Volume of nucleus}}\text{or}\ \rho\text{nu}$
$=\frac{\text{MA}}{\frac{4}{3}\pi\text{R}_0^3\text{A}}=\frac{3\text{m}}{4\pi\text{R}_0^3}$
Clearly, nuclear density is independent of mass number A or the size of the nucleus. The nuclear mass density is of the order 1017kg m-3 This density is very large as compared to the density of ordinary matter, say water, for which
$\rho$ = 1.0 × 103kg m-3