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. $\sqrt{\pi r}$

Answer 1

The correct option is B $2\pi r$

Bohr's postulate for angular momentum states,

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

For ground state, momentum of electron$=p=mv=\frac{h}{2\pi r}$

Thus de-broglie wavelength for the electron is $\frac{h}{p}=2\pi r$