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)

Angular momentum $ = \frac{nh}{2\pi}$
$\Rightarrow $ moment of momentum $= p\times r_n$
$\Rightarrow p\times r_n =\frac{nh}{2\pi}$
$\Rightarrow \frac{h}{\lambda}r_n = \frac{nh}{2\pi}$
$\Rightarrow \lambda = \frac{2\pi r_n}{n}$
For $1^{st}$ orbit, $n =1, \lambda = 2\pi r_1$
$\Rightarrow \lambda =$ circumference of $1^{st}$ orbit.