The circumference of the Bohr orbit for the H atom is related to the de Broglie wavelength associated with the electron revolving around the orbit by the following relation: 
1. $2\mathrm{\pi r}=\mathrm{n\lambda }$
2. $\mathrm{\pi r}$ $=$ $2\mathrm{n\lambda }$
3. $\mathrm{mvr}$ $=$ $\mathrm{n\lambda }$
4. $\mathrm{vr}$ $=$ $2\mathrm{n\lambda }$