FAD1022 CQ — Atomic & Nuclear Physics Quiz Set

Course quiz problems covering hydrogen energy levels, radioactive decay, carbon dating, beta decay energy, and binding energy calculations.

[!info] Quiz Set Contents This collection contains 8 problems from course quizzes covering:

  • Hydrogen atom transitions and energy levels
  • Carbon-14 dating calculations
  • Beta decay energy (Q-value)
  • Nuclear binding energy

[!tip] Quick Answer Key (8 Problems)

Q Answer Topic
1 $n=4$ Hydrogen transition to ground state
2 $0.745$ MeV Beta decay (Th-234)
3 $11,460$ years C-14 dating (25% remaining)
4 $27.4$ MeV Binding energy (He-4)
5 $2,378$ years C-14 dating (75% remaining)
6 $0.156$ MeV Beta decay (C-14)
7 $n=1$ Hydrogen transition (third excited state)
8 $37.91$ MeV Binding energy (Li-7)

Question 1: Hydrogen Transition to Ground State

Problem Statement: An electron in a hydrogen atom transitions from an excited state to the ground state and emits a photon with energy $12.75$ eV. Determine the initial energy level of the electron.

Options: $n=5$, $n=4$, $n=3$, $n=2$

Solution:

Step 1: Recall hydrogen energy level formula (Bohr Model) $$E_n = -\frac{13.6}{n^2}\ \text{eV}$$

Ground state energy: $E_1 = -13.6$ eV

Step 2: Find initial energy level energy $$\Delta E = E_n - E_1 = 12.75\ \text{eV}$$

$$E_n = E_1 + 12.75 = -13.6 + 12.75 = -0.85\ \text{eV}$$

Step 3: Solve for $n$ $$-\frac{13.6}{n^2} = -0.85$$

$$n^2 = \frac{13.6}{0.85} = 16$$

$$n = 4$$

Answer: $n=4$

[!tip] Verification $E_4 = -13.6/16 = -0.85$ eV $\Delta E = -0.85 - (-13.6) = 12.75$ eV ✓

Question 2: Beta Decay Energy (Thorium-234)

Problem Statement: Calculate the energy released in a beta decay: $$^{234}{90}\text{Th} \rightarrow {}^{234}{91}\text{X} + {}^{0}_{-1}e$$

Given:

  • Mass of Th = $234.0441$ u
  • Mass of X = $234.0433$ u

Options: $1.68$ MeV, $0.156$ MeV, $0.745$ MeV, $0.08$ MeV

Solution:

Step 1: Calculate mass difference (mass defect) $$\Delta m = m_{\text{Th}} - m_{\text{X}} = 234.0441 - 234.0433 = 0.0008\ \text{u}$$

[!note] In beta decay, the electron mass is already accounted for in the atomic mass of the daughter nucleus (X has one more proton but same mass number).

Step 2: Convert to energy using $E = \Delta m \times 931.5$ MeV/u $$Q = 0.0008 \times 931.5 = 0.7452\ \text{MeV}$$

Answer: $0.745$ MeV

Question 3: Carbon-14 Dating (25% Remaining)

Problem Statement: A piece of ancient wood is found, and analysis shows it contains $25%$ of the original Carbon-14 found in living tissue. If the half-life of Carbon-14 is $5730$ years, estimate the age of the wood.

Options: $11460$ years, $2378$ years, $1210$ years, $5378$ years

Solution:

Step 1: Recognize the fraction remaining $$\frac{N}{N_0} = 25% = \frac{1}{4} = \left(\frac{1}{2}\right)^2$$

Step 2: Determine number of half-lives Since $1/4 = (1/2)^2$, exactly 2 half-lives have passed.

Step 3: Calculate age $$t = 2 \times T_{1/2} = 2 \times 5730 = 11,460\ \text{years}$$

Answer: $11,460$ years

[!tip] Alternative Method (using decay equation) $$N = N_0 e^{-\lambda t} \Rightarrow 0.25 = e^{-\lambda t}$$ $$\lambda = \frac{\ln 2}{5730} = 1.21 \times 10^{-4}\ \text{yr}^{-1}$$ $$t = -\frac{\ln(0.25)}{\lambda} = \frac{1.386}{1.21 \times 10^{-4}} = 11,460\ \text{years}$$

Question 4: Binding Energy (Helium-4)

Problem Statement: Calculate the binding energy of an element with:

  • Number of Protons, $Z = 2$
  • Mass Number, $A = 4$

Given:

  • Mass of proton = $1.0073$ u
  • Mass of neutron = $1.0087$ u
  • Mass of nucleus = $4.0026$ u
  • $1$ u = $931.5$ MeV

Options: $20.3$ MeV, $24.7$ MeV, $27.4$ MeV, $32.6$ MeV

Solution:

Step 1: Calculate mass of constituent nucleons

  • Protons: $2 \times 1.0073 = 2.0146$ u
  • Neutrons: $2 \times 1.0087 = 2.0174$ u
  • Total: $4.0320$ u

Step 2: Calculate mass defect $$\Delta m = \text{mass of constituents} - \text{mass of nucleus}$$ $$\Delta m = 4.0320 - 4.0026 = 0.0294\ \text{u}$$

Step 3: Calculate binding energy $$E_B = \Delta m \times 931.5\ \text{MeV/u}$$ $$E_B = 0.0294 \times 931.5 = 27.386\ \text{MeV}$$

Answer: $27.4$ MeV

Question 5: Carbon-14 Dating (75% Remaining)

Problem Statement: A piece of ancient wood is found, and analysis shows it contains $75%$ of the original Carbon-14 found in living tissue. If the half-life of Carbon-14 is $5730$ years, estimate the age of the wood.

Options: $5378$ years, $11460$ years, $2378$ years, $1210$ years

Solution:

Step 1: Use radioactive decay equation $$N = N_0 e^{-\lambda t}$$

$$0.75 = e^{-\lambda t}$$

Step 2: Calculate decay constant $$\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{5730} = 1.21 \times 10^{-4}\ \text{yr}^{-1}$$

Step 3: Solve for time $$\ln(0.75) = -\lambda t$$

$$-0.2877 = -(1.21 \times 10^{-4})t$$

$$t = \frac{0.2877}{1.21 \times 10^{-4}} = 2,378\ \text{years}$$

Answer: $2,378$ years

Question 6: Beta Decay Energy (Carbon-14)

Problem Statement: Calculate the energy released in a beta decay: $$^{14}{6}\text{C} \rightarrow {}^{14}{7}\text{N} + {}^{0}_{-1}e$$

Given:

  • Mass of C = $14.003242$ u
  • Mass of N = $14.003074$ u

Options: $1.56$ MeV, $0.168$ MeV, $1.68$ MeV, $0.156$ MeV

Solution:

Step 1: Calculate mass difference $$\Delta m = m_{\text{C}} - m_{\text{N}} = 14.003242 - 14.003074 = 0.000168\ \text{u}$$

Step 2: Convert to energy $$Q = \Delta m \times 931.5\ \text{MeV/u}$$ $$Q = 0.000168 \times 931.5 = 0.1565\ \text{MeV}$$

Answer: $0.156$ MeV

Question 7: Hydrogen Transition (Third Excited State)

Problem Statement: An electron in a hydrogen atom transitions from the third excited state to a lower state and emits a photon with energy $12.75$ eV. Determine the final energy level of the electron.

Options: $n=5$, $n=4$, $n=3$, $n=2$, $n=1$, $n=\infty$

Solution:

Step 1: Identify initial state Third excited state corresponds to $n_i = 4$ (ground state is $n=1$, first excited is $n=2$, etc.)

Energy at $n=4$: $$E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85\ \text{eV}$$

Step 2: Find final energy level $$\Delta E = E_i - E_f = 12.75\ \text{eV}$$

$$E_f = E_i - 12.75 = -0.85 - 12.75 = -13.6\ \text{eV}$$

Step 3: Identify $n_f$ $$E_n = -\frac{13.6}{n^2} = -13.6$$

$$n^2 = 1 \Rightarrow n = 1$$

Answer: $n=1$ (ground state)

Question 8: Binding Energy (Lithium-7)

Problem Statement: Calculate the binding energy of an element with:

  • Proton number $Z = 3$
  • Total nucleons $A = 7$

Given:

  • Proton = $1.0073$ u
  • Neutron = $1.0087$ u
  • Mass of Nucleus: $7.0160$ u

Options: BE=$37.91$ eV, BE=$2856$ eV, BE=$36.61$ eV, BE=$2851.5$ eV

Solution:

Step 1: Calculate number of neutrons $$N = A - Z = 7 - 3 = 4$$

Step 2: Calculate mass of constituents

  • Protons: $3 \times 1.0073 = 3.0219$ u
  • Neutrons: $4 \times 1.0087 = 4.0348$ u
  • Total: $7.0567$ u

Step 3: Calculate mass defect $$\Delta m = 7.0567 - 7.0160 = 0.0407\ \text{u}$$

Step 4: Calculate binding energy $$E_B = 0.0407 \times 931.5 = 37.91\ \text{MeV}$$

Answer: BE = $37.91$ MeV (note: options say "eV" but should be "MeV")

[!warning] Unit Discrepancy The correct answer is $37.91$ MeV, not eV. This is approximately $37,910,000$ eV. The first option has the correct numerical value but incorrect unit labeling.

Summary Table

Q Problem Type Key Formula Answer
1 Hydrogen transition to ground $E_n = -13.6/n^2$ $n_i = 4$
2 Beta decay (Th-234) $Q = \Delta m \times 931.5$ $0.745$ MeV
3 C-14 dating (25%) $t = 2 \times T_{1/2}$ $11,460$ years
4 Binding energy (He-4) $E_B = [Zm_p + Nm_n - m_{\text{nucleus}}]c^2$ $27.4$ MeV
5 C-14 dating (75%) $t = -\ln(N/N_0)/\lambda$ $2,378$ years
6 Beta decay (C-14) $Q = \Delta m \times 931.5$ $0.156$ MeV
7 Hydrogen transition (n=4 → ?) $\Delta E = E_i - E_f$ $n_f = 1$
8 Binding energy (Li-7) $E_B = \Delta m \times 931.5$ $37.91$ MeV

Key Constants

Constant Value
Hydrogen ground state energy $-13.6$ eV
C-14 half-life $5730$ years
Energy equivalent of 1 u $931.5$ MeV
Mass of proton $1.0073$ u
Mass of neutron $1.0087$ u

Key Formulas

Radioactive Decay: $$N(t) = N_0 e^{-\lambda t}$$ $$\lambda = \frac{\ln 2}{T_{1/2}}$$

Beta Decay Q-value: $$Q = (m_{\text{parent}} - m_{\text{daughter}})c^2$$

Binding Energy: $$E_B = [Zm_p + Nm_n - m_{\text{nucleus}}]c^2$$ $$E_B = \Delta m \times 931.5\ \text{MeV/u}$$

Hydrogen Energy Levels: $$E_n = -\frac{13.6}{n^2}\ \text{eV}$$ $$\Delta E = E_i - E_f = hf$$

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