Question

If the radius of first Bohr’s orbit is x, then de-Broglie wavelength of electron in 3rd orbit is nearly :

• A. 2$\pi x$
• B. 6$\pi x$
• C. 9x
• D. $\frac{x}{3}$

Answer 1

The correct option is B 6$\pi x$


Radius of nth orbit is $r\infty n^2\frac{r_2}{r_1}={\left(\frac{n_2}{n_1}\right)}^2\frac{r_2}{x}={\left(\frac{3}{1}\right)}^2{r}_2=9x2\pi \left(9x\right)=n\lambda \therefore \lambda =\frac{2\pi \left(9x\right)}{3}\lambda =6\pi x$