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.