The relation between the circumference of an electron orbit in a hydrogen atom and the de broglie wavelength of the electron in the same orbit is given by

Options

A. `2pir=nlamda`
B. `2pir=(nh)/(2)`
C. `2pir = 2n lamda`
D. `2pir= (nlamda)/(4)`

Answer

A

Explanation

For the circular orbit of an electron `mvr = (nh)/(2pi)`
`therefore 2pir = n((h)/(mv)) = n lamda`
`therefore lamda = (h)/(mv) = ` de broglie wavelength
`because` Circumference `=nlamda`