Question

The radius of fourth orbit in ${He}^{+}$ ion?

Answer 1

$r=\frac{n^2{h}^2}{4{\pi }^{2}mKZ{e}^2}$

here r is the radius of nth orbit .

where h is plank's constant

m = mass of electrons

K = kappa's constant

e = charge of electron

$r=\frac{\left(0.592\right)\left(4^2\right)}{2}$

$=0.592\times 8$

$4.232A^{\circ }$