Question

For the Bohr's first orbit of circumference $2\pi r$, the de-Broglie wavelength of revolving electron will be.

• A. $2\pi r$
• B. $\pi r$
• C. $\frac{1}{2\pi r}$
• D. $\frac{1}{4\pi r}$

Answer 1

Answer: (A)

$2\pi r$

Answer 2

$mvr = \frac{nh}{2\pi}$ According to Bohr’s theory

⇒ $2\pi r = n\left( \frac{h}{mv} \right) = n\lambda$ for $n = 1$, $\lambda = 2\pi r$