Question

The ratio of areas of the electron orbits for the first excited state and the ground state for the hydrogen atom is

• A. $4:1$
• B. $16:1$
• C. $8:1$
• D. $2:1$

Answer 1

Answer: (B)

The radius of the orbit of the electron in the $n$ th excited state
$r_e=$ $\frac{n^{2}4\pi {ϵ}_0{h}^2}{4{\pi }^{2}mZ{e}^2}$
For the first excited state
$n=2,Z=1$
$\because r^{{}^{\prime }}=$ $\frac{4{ϵ}_0{h}^2}{\pi m{e}^2}$
For the ground state of hydrogen atom
$n=1,Z=1$
$\because r^{{}^{\prime \prime }}=$ $\frac{h^2{ϵ}_0}{\pi m{e}^2}$
The ratio of radius
$\frac{r^{{}^{\prime }}}{r{}^{{}^{\prime \prime }}}=\frac{4}{1}$
The ratio of area of the electron orbit for hydrogen atom
$\frac{{\left(A\right)}^{{}^{\prime }}}{{\left(A\right)}^{{}^{\prime \prime }}}=\frac{4\pi {\left(r^{\prime }\right)}^2}{4\pi {\left(r^{\prime \prime }\right)}^2}$
$\frac{A^{{}^{\prime }}}{A{}^{{}^{\prime \prime }}}=\frac{16}{1}$