The de broglie wavelength of an electron in the first Bohr orbit is equal to

Options

A. Diameter of the first orbit
B. Circumference of the first orbit
C. Squareroot of the area of the first orbit
D. Twice the circumference of the first orbit

Answer

B

Explanation

`lamda = (h)/(mv)` and for the first orbit , `mvr=(h)/(2pi)`
`therefore (h)/(mv) = 2pir`
`therefore lamda =` Circumference