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

From Bohr's law
`mvr=(nh)/(2pi)=(3h)/(2pi)` (from ques.)
`implies n=3`
and momentum `=mv=(3h)/(2pi r)`
Now, radius of `n^(th)` shell, `r=(n^(2)/z)a_(0)`
`implies r=((3)^(2))/3.a_(0)" "[ :' Z_(Li)=3]`
`implies r=3a_(0)`
from De broglie law
wavelength `=(h)/("momentum")`
`implies lambda=(h)/(mv)=h/((3h)/(2pi) r)`
`implies lambda=(2pi r)/3=(2pi)/3xx3a_(0)`
`lambda=2 pia_(0)=p pia_(0)`
`implies P=2`