The de Broglie wavelength of an electron in the first Bohr orbit is

Options

A. equal to halff the circumference of the first orbit
B. equal toone fourth the circumference of the first orbit
C. equal to the circumference of the first orbit
D. equal to twice the circumference of the first orbit.

Answer

C

Explanation

The de Broglie wavelength of an electron is given by: `lamda=h//mv` . . (i)
From Bohr.s postulates of atomic structure we have
`mvr=(nh)/(2pi)`
(Angular momentum=integral multiple of `h//2pi`)
`therefore`for the first Bohr orbit, n=1
and `mvr=(h)/(2pi) or mv=(h)/(2pir)` . (ii)
From (i) and (ii) we get
`lamda=(h)/(h//2pir)=2pir` (circumference of first orbit).