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

Angular momentum of electron is given as follows :
`l=(3h)/(2pi)`
We can understand that electron is in quantum state n=3
According to de Broglie.s hypothesis, circumference of the orbital must be integral multiple of wavelength .
`nlambda=2pir_n`
`rArr nlambda=2pi n^2/Z a_0`
`rArr lambda=2pi n^2/Z a_0`
We can substitute n=3 , Z=3 to get the de Broglie wavelength as follows :
`lambda=2pi(3/3)a_0=2pia_0`
Now comparing it with given result `ppia_0` we get p=2 .