Question

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

Answer 1

Use mvr=nh/2π

=> mvr=nh/2π
=> 2πr=nh/mv
=> 2πr=n×lambda {lambda=h/mv}

now, n=4 and wavelength=4A

=> 2πr(circumference) = 4×4
= 16A

Answer 2

Answer:

16

Explanation:

De Broglie wavelength of Hydrogen atom = 4A (Given)

A De Broglie wavelength is the wavelength, that is manifested in all the particles in quantum mechanics. According to the duality of the wave-particle, it determines the probability density of finding the object at a given point of the configuration space.

Using the equation - mvr=nh/2π

= mvr =nh/2π

= 2πr = nh/mv

= 2πr = n× λ { Since λ =h/mv}

Since, n=4 and wavelength =4A

Thus,

= 2πr = 4×4

= 16A

Therefore, the circumference of the orbit is 16.