Question

The radius of the first Bohrs orbit in H-atoms is $r_1$. The corresponding wavelength of an electron in 2nd orbit is:

• A. $6\pi r_1$
• B. $4\pi r_1$
• C. $2\pi r_1$
• D. $3\pi r_1$

Answer 1

Answer: (B)

$4\pi r_1$

$\frac{m{v}^2}{r}=\frac{z{e}^2}{r^2}$

$r=\frac{m^2{v}^2{r}^2}{z{e}^{2}m}=\frac{(mvr{)}^2}{z{e}^{2}m}$

$mvr=\frac{nh}{2\pi }$

$mv{r}_1=\frac{n}{2\pi }=h$

$mv{r}_2=2h$

$r_1=\frac{h^2}{z{e}^{2}m}$

$r_2=\frac{4h^2}{z{e}^{2}m}$

$\Rightarrow r_2=4r_1$

So wave length corresponding to $2^{nd}$ Bohr orbit $=\frac{2\pi r_2}{n}$

$n=2$

$=4\pi r_1$