An electron is an excited state of `Li^(2 + )`ion has angular momentum `3h//2 pi ` . The de Broglie wavelength of the electron in this state is `p pi a_(0) (where a_(0) ` is the bohr radius ) The value of p is

Options

A. `4`
B. `3`
C. `2`
D. `1`

Answer

C

Explanation

Given angular momentum of electron,
`L=(3h)/(2pi)=(nh)/(2pi)`
It means, the electron is in quantum state, n=3.
form de-Broglie hypothesis
`nlambda=2pir_(n)=2pi((n^(2))/Za_(0)) [:' r_(n)=(a_(0)n^(2))/Z]`
or `lambda=2pi n/Z a_(0)`
for lithium, Z=3 and n=3, so
`lambda=2pi 3/3a_(0)=2pia_(0)=p pia_(0)` (given)
So P=2