Question

If the de-Broglie wave length of the fourth Bohr's orbit of hydrogen atom is 4
, the circumference of the orbit is?

Answer 1

we know that wavelength equal to n multiple of 2πr
then
4 =4×2πr
then 2πr=1 hope it help u

Answer 2

Answer : The circumference of the orbit is, 16

Solution : Given,

wavelength = 4

n = 4

According to the De-Broglie, the relation between the wavelength and momentum will be,

$\lambda=\frac{h}{mv}$     ..........(1)

where,

$\lambda$ = wavelength

h = Planck's constant

m = mass

v = velocity

According to the De-Broglie postulates, the angular momentum must be a integral multiple of its wavelength.

$mvr=\frac{nh}{2\pi}$       ............(2)

r = radius of orbit

n = an integer

Now put the value of wavelength from equation 1 to equation 2, we get

$2\pi r=n\lambda$

Now put all the given values in this relation, we get the circumference of the orbit.

$2\pi r=4\times 4=16$

Therefore, the circumference of the orbit is, 16