Question

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

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

Answer 1

Answer: (B)

Perimeter of the orbit

Answer 2

The de-Brogle’s wavelength λ = $\frac{h}{mv}$ …(1)

Where the angular momentum of an orbiting electron is given as on

mvr = $\frac{nh}{2\pi}$ for first orbit n = 1

⇒ mv = $\frac{h}{2\pi r}$ …(2)

Using (1) and (2) we obtain

λ = $\frac{h}{\left( \frac{h}{2\pi r} \right)}$ = 2πr where, 2πr = perimeter of the orbit