Question

What would be the radius of 2nd excited state in Li^+2 ion?

Answer 1

hey!!

the formula for radius of atom = 0.529 * n^2/Z

So, here the n is 2 & Z is 3

Radius=0.529 * 4/3

            =0.705

hope this helps you...!

Answer 2

Answer: The radius of 2nd excited state in $Li^{2+}$ ion is 70.53 pm.

Explanation:

To calculate the radius of the excited state, we use the equation given by Bohr fo $n^{th}$ orbit is:

$r_n=\frac{n^2\times 52.9}{Z}$

where,

$r_n$ = radius of $n^{th}$ orbit  calculated it pm

n = number of orbit = 2

Z = atomic number = 3 (for lithium)

Putting values in above equation, we get:

$r_2=\frac{(2)^2\times 52.9}{3}\\\\r_n=70.53pm$

Hence, the radius of 2nd excited state in $Li^{2+}$ ion is 70.53 pm.