Question

Using Coulomb's law, given Li 2+ and Be 3+ (Be 3+ having the lower energy). If the electron to nucleus distance is the same, by what factor do two energies differ? (Round two places after the decimal point)

Answer 1

Answer on Question #54467 – Chemistry – General chemistry

Question:

Using Coulomb's law, given Li $2+$ and Be $3+$ (Be $3+$ having the lower energy). If the electron to nucleus distance is the same, by what factor do two energies differ? (Round two places after the decimal point)

Solution:

The electrostatic potential energy, $E_{\mathrm{elec}}$, is given by

$$E = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r}$$

where

$\epsilon_0$ is the permittivity of the vacuum,

$Z$ is the atomic number (number of protons in the nucleus),

$e$ is the elementary charge (charge of an electron),

$r$ is the distance of the electron from the nucleus.

For Li $Z_{\mathrm{Li}} = 3$, for Be $Z_{\mathrm{Be}} = 4$.

Thus,

$$\frac{E_{Li}}{E_{Be}} = \frac{Z_{Li}}{Z_{Be}} = \frac{3}{4} = 0.75$$ $$\frac{E_{Be}}{E_{Li}} = \frac{Z_{Be}}{Z_{Li}} = \frac{4}{3} = 1.33$$

Answer: Energy for Be$^{3+}$ is larger than that of Li$^{2+}$ by a factor of 1.33

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