Question

Show the circumference of the Bohr’s orbit for the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

Answer 1

Hint: According to classical mechanics based on Newton’s Laws of motion describing the motion of all macroscopic objects has a particle-like behavior. But this law fails when applied to microscopic objects like electrons, molecules, atoms, etc. the dual behavior of matter explained by de Broglie’s wavelength which means the photon has wavelength as well as momentum.

Complete step by step solution:
According to Bohr’s model for a hydrogen atom, the electron in the hydrogen atom can move around the nucleus in a circular path of fixed radius and energy, which paths are known as orbits, or stationary states, or allowed energy states. These orbitals are arranged around the nucleus based on their energy.
The expression for the angular momentum is $mvr=\dfrac{nh}{2\pi }$
$2\pi r=\dfrac{nh}{mv}$ --- (i)
From the de Broglie’s equation of the wavelength,
$\lambda =\dfrac{h}{mv}$ -- (ii)
Where m = mass of particle, v = velocity and, h = planck's constant
Equation (i) and (ii),
Then $2\pi r=n\lambda $
Hence, from the above equation the circumference of the Bohr’s orbit for the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

Note: When an electron undergoes diffraction confirmed experimentally the prediction of de Broglie’s prediction of dual behavior of matter. According to de Broglie’s, every object in motion has a wave character. The wavelengths associated with ordinary matter are short and their wave properties cannot be detected.