Question 1

Calculate the Coulombic potential energy of an electron at a distance of 52.9 pm from a lithium, Li²⁺, nucleus. A) -4.37 *10^-18 J B) -1.31 *10^-17 J C) -5.20 *10²⁰ J D) 9.63* 10⁶ J

Accepted Answer

The Coulombic potential energy is calculated using the formula $ U = \frac{k \cdot q_1 \cdot q_2}{r} $, where $ k $ is Coulomb's constant ($ 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 $), $ q_1 $ and $ q_2 $ are the charges of the electron and the Li$^{2+}$ nucleus, and $ r $ is the distance in meters. Substituting the values, the potential energy is approximately $-1.31 \times 10^{-17} \, \text{J}$.

Question 2

For a hydrogen atom, calculate the energy of an electron with principal quantum number n = 3. A) -2.421 *10^-19 J B) -1.835 *10^-23 J C) -2.421 *10^-17 J D) -1.835 *10^-25 J

Accepted Answer

The energy levels of a hydrogen atom can be calculated using the formula $E_n = -\frac{13.6 \text{ eV}}{n^2}$, where $n$ is the principal quantum number. For $n = 3$, $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.511 \text{ eV}$. Converting this energy to joules (using $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$), we get $E_3 = -1.511 \times 1.602 \times 10^{-19} \text{ J} = -2.421 \times 10^{-19} \text{ J}$, which matches option A.

Question 3

Calculate the wavenumber of the lowest energy transition in the B series of the electronic spectrum of the He⁺ ion. The Rydberg constant for He is R_He = 109707 cm^-1. A) 109707 cm^-1 B) 5332 cm^-1 C) 21941 cm^-1 D) 21332 cm^-1

Accepted Answer

The lowest energy transition in the B series of the electronic spectrum of the He⁺ ion is from the n = 3 to the n = 2 level. The wavenumber of this transition is given by the Rydberg formula:$$ ilde{v} = R_H(\frac{1}{n_1^2} - \frac{1}{n_2^2})$$where R_H is the Rydberg constant for hydrogen (109677 cm^-1) and n_1 and n_2 are the principal quantum numbers of the lower and upper levels, respectively.Plugging in the values for n_1 and n_2, we get:$$ ilde{v} = 109677 cm^{-1}(\frac{1}{2^2} - \frac{1}{3^2}) = 21332 cm^{-1}$$

Question 4

A line in the Balmer series of the emission spectrum of atomic hydrogen is observed at a wavelength of 486.3 nm. Deduce the upper state principal quantum number for this transition.&#10;A) 2&#10;B) 3&#10;C) 4&#10;D) 5

Accepted Answer

The Balmer series in the hydrogen spectrum corresponds to transitions from higher energy levels to the n=2 level. The wavelength of 486.3 nm is specifically associated with the transition from the n=4 level to the n=2 level.

Question 5

The Rydberg constant for the Be³⁺ ion is 109731 cm^-1. Calculate the ionization energy of a ground-state Be³⁺ ion. A) 218 eV B) 13.6 eV C) 54.4 eV D) 9.3 eV

Accepted Answer

The ionization energy for a hydrogen-like ion is given by the formula E = Z² * 13.6 eV, where Z is the atomic number. For Be³⁺, Z = 4, so E = 4² * 13.6 eV = 16 * 13.6 eV = 217.6 eV, which is approximately 218 eV.

Question 6

Calculate the overall change in energy when an electron is transferred from a sodium atom, Na, to a chlorine atom, Cl, to form a sodium ion, Na⁺, and chlorine ion, Cl^-. The first ionization energy of sodium is 5.14 eV. The electron affinity of chlorine is +3.62 eV.

A) 1.52 kJ mol^-1
B) 845 kJ mol^-1
C) 147 kJ mol^-1
D) 495 kJ mol^-1

Accepted Answer

The overall change in energy is the sum of the first ionization energy of sodium and the electron affinity of chlorine. The first ionization energy of sodium is 5.14 eV, and the electron affinity of chlorine is +3.62 eV. Therefore, the overall change in energy is 5.14 eV + 3.62 eV = 8.76 eV. Converting to kJ/mol, we get 8.76 eV * (1.602 x 10^-19 J/eV) * (1 kJ/1000 J) * (1 mol/6.022 x 10^23 atoms) = 147 kJ/mol.

Question 7

To which orbitals may 5d electrons make spectroscopic transitions in a hydrogenic atom?&#10;A) ns and np&#10;B) np and nd&#10;C) ns and nd&#10;D) np and nf

Accepted Answer

In a hydrogenic atom, electrons can transition between orbitals with different angular momentum quantum numbers (l). For 5d electrons (l=2), allowed transitions are to orbitals with l=1 (np) and l=3 (nf).

Question 8

Which terms can arise from the electronic configuration [He]2s²2p²? A) ³D, ¹D, ³P, ¹P, ³S and ¹S B) ¹D, ³P, and ¹S C) ³D, ¹D D) ¹S

Accepted Answer

The answer of Which terms can arise from the electronic...

Question 9

Atomic sulfur, S, has a ground electronic configuration that gives rise to the levels ³P₂_,1,0, ¹D₂ and ¹S₀. Use Hund's rule to predict which of these levels is lowest in energy. A) ¹S₀ B) ¹D₂ C) ³P₀ D) ³P₂

Accepted Answer

Hund's rule states that for a given electron configuration, the term with the highest multiplicity (highest spin) lies lowest in energy. Among the given options, D) ³P₂ has the highest multiplicity, indicating it is the lowest in energy.

Deck 8: Atomic Structure