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

Answer

`2`

Explanation

`L=(nh)/(2pi)=(3h)/(2pi)`
`n=3`
`lambda=(h)/(p)=(h)/(mv)=(h2pir)/(3h)=(2pir)/(3)`
`r=a_(0)(n^(2))/(Z)`
`lambda=(2pi)/(3)a_(0)(n^(2))/(Z)=(2pi)/(3)a_(0)(3^(2))/(3)=2pia_(0)`