Question

The de-Broglie wavelength of an electron in an orbit of circumference $2\pi r$ is

• A. $2\pi r$
• B. $\pi r$
• C. $1/2\pi r$
• D. $1/4\pi r$

Answer 1

Answer: (A)

$2\pi r$

Answer 2

According to Bohr's theory $mvr = n\frac{h}{2\pi}$

⇒ $2\pi r = n\left( \frac{h}{mv} \right) = n\lambda$

For n = 1 λ = 2πr