Question

The de Brogile wavelength of electron in first Bohr orbit is exactly equal to ?

• A. Half the circumference of the orbit
• B. Equal to the circumference of the orbit
• C. Twice the circumference of the orbit
• D. Thrice the circumference of the orbit

Answer 1

The correct option is B

Equal to the circumference of the orbit

Angular momentum $mvr=n\frac{h}{2\pi }=\frac{h}{2\pi }$ (For first orbit)
or, $mv=\frac{h}{2\pi r}=\frac{h}{circumference}=\frac{h}{\lambda }$ $(\lambda =\frac{h}{mv})$
Hence, $\lambda$ = circumference