Physics - Modern Physics Question with Solution | TestHub

The radius of the orbit of the electron in the$n$th excited state

$r_e =$$\frac{n^{2}4\pi {\epsilon }_0{h}^2}{{4\pi }^2{mZe}^2}$

For the first excited state

$n=2 , Z=1$

$\because r^{\prime }=$$\frac{{4\epsilon }_0{h}^2}{{\pi me}^2}$

For the ground state of hydrogen atom

$n=1 , Z=1$

$\because r^{″}=$$\frac{h^2{\epsilon }_0}{{\pi me}^2}$

The ratio of radius

$\frac{r^{\prime }}{r^{″}}=\frac{4}{1}$

The ratio of area of the electron orbit for hydrogen atom

$\frac{A^{\prime }}{A^{″}}=\frac{4\pi {\left(r\text{'}\right)}^2}{4\pi {\left(r\text{'}\text{'}\right)}^2}$

$\frac{A^{\prime }}{A^{″}}=\frac{16}{1}$