Final answer:
The energy of an electron in the n = 3 state of a Be3+ ion can be calculated using a modified Bohr model equation, which considers the atomic number and principal quantum number. By plugging in the given values, the computed energy for this state is -3.874 x 10^-18 joules.
Step-by-step explanation:
Calculating the Energy of an Electron in a Be3+ Ion
To calculate the energy of an electron in the n = 3 state of a Be3+ ion, we can use the modified Bohr model formula specifically for hydrogen-like ions, which is given by:
E = -µZ2e4 / (8 ε02h2n2)
where µ is the reduced mass of the electron, Z is the atomic number of the ion, e is the elementary charge, ε0 is the vacuum permittivity, h is the Planck's constant, and n is the principal quantum number.
In this specific example, since we are dealing with a Be3+ ion, which is similar to a hydrogen atom but with a higher nuclear charge, we must adjust the formula to account for the increased charge. The modified equation becomes:
E = - Z2 k / n2
Plugging in the values given (Z = 4 for beryllium, k = 2.179 x 10-18 J, and n = 3), we get:
E = - 42 × (2.179 x 10-18 J) / 32
E = - (16 × 2.179 x 10-18 J) / 9
E = - (34.864 x 10-18 J) / 9
E = -3.874 x 10-18 J
Therefore, the calculated energy of the electron in the n = 3 state of a Be3+ ion is -3.874 x 10-18 joules.