Question

Fill in the Blank: An electron in an excited state of $Li^{2+}$ ion has angular momentum $3h/2\pi$.The de-Broglie wavelength of the electron in this state is pπa0 (where a0 is the Bohr radius). The value of p is . . . . . . .

Answer 1

i think 48 but not sure if right then mark Brainliest

Answer 2

Answer: here ang momentum, L=nh/2π=3h/2π

So n=3

We know,

n£= 2πr

£=2πr/n

£=pπa°=2πr/n

P=2r/na°

We have r=a°n^2/Z , n=3 Z=3

So P=2*n/Z

P=3

Hope it helps

Explanation: