Question

If the radius of first Bohr orbit is $x$, then de Broglie wavelength of electron in $3^{rd}$ orbit is nearly.

• A. $2\pi x$
• B. $6\pi x$
• C. $9x$
• D. $x/3$

Answer 1

The correct option is B $6\pi x$

Radius of 3rd orbit radius $=9x=n^{2}x$ (where $n=3$ )
Let de broglie wavelength be $\lambda$.
For the interference of the waves to be constructive,
$n\lambda =2\pi r$ ($r$ is radius of orbit)
$\Rightarrow \lambda =\frac{2\pi \times 9x}{3}$ (where, $n=3$, the quantum state)
$\Rightarrow \lambda =6\pi x$