Question

The radius of first Bohr's 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

The correct option is B

$4\pi r_1$

From Bohr's postulates, we know

Wavelength = $\frac{circumference}{Numberofwaves}$

Number of waves in nth orbit = $n$

Therefore,

$\lambda$ = $\frac{2\pi r_n}{n}$

& we know

$r_n$ = $r_1\left(n^2\right)$

$r_2$ = $r_1\left(2^2\right)$

= $4r_1$

$\Rightarrow$ $\lambda$ = $\frac{2\pi \times 4r_1}{2}$

$\Rightarrow$ $\lambda$ = $4\pi r_1$