If force of attraction  between the electron and nucleus in 2nd orbit of Li2+ is f, what is the force of attraction if electron present in 1st orbit of H ?

Dear Student,

Please find below the solution to the asked query.

The formula for force of attraction between the electron and the nucleus is:
$F = \frac{KZ{e}^2}{r^2}$
where F = Force
K = constant
e = charge (constatnt)
Z = atomic number
r = radius of the orbit

And,
$r_n = \frac{a_o{n}^2}{Z}$
where a_o = constant
n = number of orbit.

Thus we get
$F = \frac{K e^2}{{a_o}^2}\frac{Z^3}{n^4}$
For Li²⁺, Z = 3 and n = 2 (2nd orbit), force is equal to 'f'
$f=\frac{K e^2}{{a_o}^2}\frac{(3{)}^3}{(2{)}^4}=\frac{K e^2}{{a_o}^2}\frac{27}{16} --------\left(1\right)$
And for hydrogen, Z =1, n = 1 (1st orbit), force will be:
$$\begin{gathered} Force = \frac{K e^2}{a_o^2}\frac{(1{)}^3}{(1{)}^2}=\frac{K e^2}{a_o^2} \\ Multiply and divide by \frac{27}{16} \\ Force =\frac{K e^2}{a_o^2}\frac{27}{16}\times \frac{16}{27} \\ = f\times \frac{16}{27} (from equation 1) \\ =\frac{16f}{27} \end{gathered}$$

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