Question

Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wave length associated with the electron revolving arround the nucleus.

Answer 1

We know that for The angular momentum of an electron ⇒

$mvr = \frac{nh}{2 \pi }$  -----→ (i)

Also, By de broglie equation , we know that 

λ = h/mv

$mv = \frac{h}{v}$  ----→(ii)

Now putting the value of (ii) to (i)

$2 \pi r = n(h/mv)$

2πr = nλ 

Since ‘2πr’ represents here, circumference of the Bohr orbit (r)

Therefore the given sentence of the question is proved.

Hope it Helps. :-)

Answer 2

Hydrogen just has one atom, the angular momentum of its electron is given by:-

$m \nu r = \dfrac{nh}{2 \pi}$

Here, 'n' represents the shell number

According the equation by De-Broglie:-

$\lambda = \dfrac{h}{m \nu}$

$m \nu = \dfrac{h}{\lambda}$

Substitute the value of $m \nu$ from here in first equation:-

$\dfrac{hr}{\lambda} = \dfrac{nh}{2 \pi}$

$2 \pi r = n \lambda$

Here, $2 \pi r$ represents the circumferene of the Bohr's orbit. It is proved that circumference of the Bohr's orbit of Hydrogen atom is integeral multiple of de-Broglie's wavelength associated with the electron.