Question

The de-Broglie wavelength of an electron in first orbit of Bohr's hydrogen atom is equal to -

• A. Radius of the orbit
• B. Perimeter of the orbit
• C. Diameter of the orbit
• D. Half of the perimeter of the orbit

Answer 1

Answer: (B)

Perimeter of the orbit

Answer 2

The de-Broglie wavelength

l = $\frac{h}{mv}$ ….(1)

where the linear momentum of orbiting electron is given as :

mvr =$\frac{nh}{2\pi}$, for first orbit, n =1, mv =$\frac{h}{2\pi r}$….(2)

Using equation (1) and (2), we get :

l = $\frac{h}{(h/2\pi r)}$= 2pr = perimeter of first orbit