Question

The de-Broglie wavelength of an electron in the first Bohr orbit is

• A. Equal to one-fourth the circumference of the first orbit
• B. Equal to half the circumference of first orbit
• C. Equal to twice the circumference of first orbit
• D. Equal to the circumference of the first orbit.

Answer 1

Answer: (D)

Equal to the circumference of the first orbit.

Answer 2

Angular momentum $= \frac{nh}{2\pi}$

$\Rightarrow$ moment of momentum $= \frac{nh}{2\pi}$

$\Rightarrow p \times r_{n} = \frac{nh}{2\pi}$

$\frac{h}{\lambda}r_{n} = \frac{nh}{2\pi} \Rightarrow \lambda = \frac{2\pi r_{n}}{n}$

For 1^st orbit, n = 1, $\lambda = 2\pi r_{1}$

$\Rightarrow \lambda =$ Circumference of 1^st orbit.