The de Broglie wave present in fifth Bohr orbit is:

Options

A.
B.
C.
D.

Answer

D

Explanation

For a stationary Bohr orbit , `mvr = (nh)/(2pi)`
`i.e. 2pir = (nh)/(mv) = n((h)/(mv)) = n lamda`
where `lamda = " de broglie wavelength " lamda = (h)/(mv)` and n is an integer.
Thus only that circular orbit can be in a stationary energy state which contains an integral number of de broglie wavelength i.e. `2pir = n lamda` . For the fifth orbit `2pir = 5 lamda`
In figure (d) there are 5 wavelength .
(a) , (b) and (c ) have `3/2 lamda, 2lamda " and " 3 lamda` respectively.
`therefore ` Figure (d) gives the de broglie wave present in the 5th Bohr orbit .