Numerical Problems Shortcuts and Tricks:

1. Calculate the nucleus radius of a gold atom (atomic number 79 and mass number 197):

• The formula for nuclear radius is: $$R = k * A^{1/3}$$ Where:
• (R) is the nuclear radius in fermi (fm).
• (A) is the mass number of the atom.
• (k) is a constant approximately equal to 1.2 fm. $$R = 1.2 \text{ fm} \times (197)^{1/3}$$ $$R \approx 7.41 \text{ fm}$$

2. Determine the number of protons, neutrons, and electrons in magnesium-24:

• The atomic number of magnesium is 12, which means it has 12 protons.
• The mass number of magnesium-24 is 24, which means it has a total of 24 nucleons (protons + neutrons).
• Subtracting the number of protons from the total nucleons gives the number of neutrons: $$24 - 12 = 12 \text{ neutrons}$$
• Since the atom is neutral, the number of electrons is equal to the number of protons: $$12 \text{ electrons}$$

3. Calculate the wavelength of the photon emitted when an electron transitions from n = 3 to n = 2 in a hydrogen atom:

• Use the Rydberg formula: $$\frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$ Where:
• (\lambda) is the wavelength of the emitted photon in meters (m).
• (R_H) is the Rydberg constant for hydrogen (approximately 1.0973731 × 10^7 m^-1).
• (n_f) is the final energy level (2).
• (n_i) is the initial energy level (3). $$ \frac{1}{\lambda} = (1.0973731 \times 10^7 \text{ m}^{-1}) \left(\frac{1}{2^2} - \frac{1}{3^2}\right)$$ $$ \lambda \approx 656.3 \text{ nm}$$

4. Determine the frequency of the light emitted when an electron transitions from n = 4 to n = 2 in a lithium atom:

• The formula for the frequency of emitted light is: $$f = \frac{c}{\lambda}$$ Where:
• (f) is the frequency in hertz (Hz).
• (c) is the speed of light (2.998 × 10^8 m/s).
• (\lambda) is the wavelength of the emitted light (in meters).
• We already calculated (\lambda) to be 656.3 nm, which is (656.3 \times 10^{-9} \text{ m}). $$ f = \frac{2.998 \times 10^8 \text{ m/s}}{656.3 \times 10^{-9} \text{ m}}$$ $$f \approx 4.570 \times 10^{14} \text{ Hz}$$

5. Calculate the energy required to remove an electron from the outermost shell of a sodium atom:

• Use the ionization energy formula: $$E = h c R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$ Where:
• (E) is the energy required for ionization in joules (J).
• (h) is Planck’s constant (6.626 × 10^-34 Js).
• (c) is the speed of light (2.998 × 10^8 m/s).
• (R_H) is the Rydberg constant for hydrogen (1.0973731 × 10^7 m^-1).
• (n_f) is the final energy level (∞, since the electron is removed from the atom).
• (n_i) is the initial energy level (1, since it’s the outermost shell). $$E = (6.626 \times 10^{-34} \text{ Js}) \times (2.998 \times 10^8 \text{ m/s}) \times (1.0973731 \times 10^7 \text{ m}^{-1}) \left(\frac{1}{\infty^2} - \frac{1}{1^2}\right)$$ $$E = 5.14 \times 10^{-19} \text{ J}$$

6. Determine the ionization energy of a helium atom, given an energy requirement of 2372 kJ/mol:

• Energy per atom is determined by: $$ E = \frac { Energy \ required }{Avogadro’s \ Number} $$ $$ E = \frac {2372 kJ/mol}{ 6.022 \times 10^{23} mol^{-1}} $$ $$ Eatom = 3.94 \times 10^{-19} \text {J} $$
• Now just solve the usual ionization energy formula for $$ R_H $$ $$ E = h c R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$ $$ R_H = \frac { E}{h c} (\frac{1}{n_f^2} - \frac{1}{n_i^2})$$ $$ R_H = \frac{3.94 \times 10^{-19} J}{(6.626 \times 10^{-34} Js) (2.998 \times 10^8 ms^{-1}) } (\frac{1}{\infty^2} - \frac{1}{1^2})$$ $$ R_h = 1.097 \times 10^7 m^{-1}$$

7. Calculate the Rydberg constant from experimental data:

• The Rydberg constant can be calculated using the formula: $$R_H = \frac{1}{\lambda_{\text{measured}}} - \frac{1}{\lambda_{\text{calculated}}}$$ Where:
• (R_H) is the Rydberg constant in meters^-1.
• (\lambda_{\text{measured}}) is the experimentally measured wavelength of the emitted photon in meters.
• (\lambda_{\text{calculated}}) is the calculated wavelength using the formula with known values.

8. Determine the atomic number of an element if the wavelength of its emitted light is 656.3 nm and the Rydberg constant is known:

• Use the Rydberg formula: $$\frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$ Where:
• (\lambda) is the wavelength of the emitted light (given as 656.3 nm or (656.3\times10^{-9} \text{m})).
• (R_H) is the Rydberg constant (given or measured).
• (n_f) is the final energy level (2).
• (n_i) is the initial energy level (unknown).

Rearrange the formula to solve for (n_i): $$n_i = \sqrt{\frac{1}{R_H \lambda} + \frac{1}{4}}$$ $$n_i = \sqrt{\frac{1}{(1.0973731 \times 10^7 \text{ m}^{-1}) (656.3\times10^{-9} \text{m})} + \frac{1}{4}}$$ $$n_i \approx 3$$

Since (n_i) represents the energy level of the electron before the transition, the atomic number of the element is 3, which corresponds to lithium (Li).

9. Calculate the mass defect of an atom of carbon-12:

• Mass defect is the difference between the actual mass and the sum of the masses of its protons and neutrons.
• Given the actual mass of (12.000 \text{ amu}) and sum of proton and neutron masses of (12.011 \text{ amu}): $$ \Delta m = 12.000 \text{ amu} - 12.011 \text{ amu} = -0.011 \text{ amu} $$
• The negative value indicates a mass defect, which means the actual mass of the carbon-12 atom is less than the sum of its individual nucleons.

10. Determine the binding energy per nucleon of an atom of uranium-238:

• Binding energy per nucleon is calculated by: $$E_b =