Question

The de-Broglie wavelength of an electron in the ground state of the hydrogen atom is

• A. $\pi r^2$
• B. $2\pi r$
• C. $\pi r$
• D. $\sqrt{2\pi r}$

Answer 1

Answer: (B)

According to Bohr's theroy
$mvr=\frac{nh}{2\pi }$
$2\pi r=\frac{nh}{mv}$
$2\pi r=n\lambda$
For n=1, $\lambda =2\pi r$