Question

The radius of first excited state of Be3+ ion???​

Answer 1

Answer:

1.587 A°

Explanation:

• formula for any radius of hydrogen like elements is r=0.529×n^2/Z A° here n=3, and Z=3. r=0.529*3 A°

Answer 2

Answer:

$r=5.292 \times 10^{-11}\; m$

Explanation:

The expression for the Bohr's radius is:

$r=\frac{n^2}{Z} \times 5.292 \times 10^{-11}\; m$

Where, n is the energy level

Z is the atomic number

First excited state, n = 2

For, $Be^{3+}$, Z = 4

So,

$r=\frac{2^2}{4} \times 5.292 \times 10^{-11}\; m$

$r=5.292 \times 10^{-11}\; m$