Chemistry: According to coulomb s law, how does the force between a hydrogen nucleus and an electron in the n =2 level compare with that for one in the n = 1 level if the distance between the nucleus and the n = 2 level is twice as great as that between the nucleus and the n = 1 level? one-quarter as strong four times as strong one-half as strong twice as strong

Velocity because it is speed in direction....

According to coulomb s law, how does the force

Ответ:

deidaralove90

11.01.2022

Chemistry

one-quarter as strong

Explanation:

Coulomb's law gives the mathematical expression to calculate the electrical force (F) between the charge of the nucleus (q₊) and the charge of an electron (q₋) separated by a distance (r).

$F=\frac{k.q_{+}.q_{-}}{r^{2} }$

where,

k is the Coulomb's constant

The force between the nucleus and an electron in the level 1 is:

$F_{1}=\frac{k.q_{+}.q_{-}}{r_{1}^{2} }$

Considering the distance to an electron from n = 2 is twice as great as the distance to an electron from n = 1, the force between the nucleus and an electron in the level 2 is:

$F_{2}=\frac{k.q_{+}.q_{-}}{r_{2}^{2} }=\frac{k.q_{+}.q_{-}}{(2r_{1})^{2} }=\frac{1}{4} \frac{k.q_{+}.q_{-}}{(r_{1})^{2} }=\frac{1}{4}F_{1}$

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