### Shortcut Methods

**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 =