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) $\dfrac{1}{{2\pi r}}$
(D) $\dfrac{1}{{4\pi r}}$

Answer 1

Hint:The de Broglie wavelength of the revolving electron can be determined by using the Bohr’s theory of the angular momentum of the electron. And then by using the given information in the question, the wavelength of the first orbit is determined.

Useful formula:
The Angular momentum of the electron is given by the Bohr’s theory,
$mvr = \dfrac{{nh}}{{2\pi }}$
Where, $m$ is the mass of the electron, $v$ is the velocity of the electron, $r$ is the radius of the orbit, $n$ is the orbit in which the electron is and $h$ is the Planck’s constant.

Complete step by step solution:
Given that,
The orbit where the electron is revolving, $n = 1$.
Now, according to the Bohr’s theory,
The Angular momentum of the electron is given by the Bohr’s theory,
$mvr = \dfrac{{nh}}{{2\pi }}\,...................\left( 1 \right)$
By taking the terms in denominator in RHS to the LHS, then the equation (1) is written as,
$2\pi mvr = nh$
By keeping the circumference $2\pi r$ in one side and the other terms in the other side, then
$2\pi r = \dfrac{{nh}}{{mv}}$
By substituting the $n$ value in the above equation, then the above equation is written as,
$2\pi r = \dfrac{h}{{mv}}$
The de Broglie wavelength is given as $\lambda = \dfrac{h}{{mv}}$, substituting this in the above equation, then the above equation is written as,
$2\pi r = \lambda $
Thus, the above equation shows the de Broglie wavelength of a revolving electron.

Hence, the option (A) is the correct answer.

Note:In the first orbit, the wavelength of the revolving electron will be equal to the circumference of the first orbit. The electron moving in its circular orbit behaves like a particle wave. The angular momentum of the electron around the nucleus is quantized.