Things Around You Are Emitting Infrared Radiation That Inclu

Things around you are emitting infrared radiation that includes the wavelength 1.03 10 m. What is the energy of these IR photons? eV Gamma rays (γ-rays) are high-energy photons. In a certain nuclear reaction, a γ-ray of energy 0.779 MeV (million electronvolts) is produced. Compute the frequency of such a photon.

Infrared (IR) radiation is a form of electromagnetic radiation with longer wavelengths compared to visible light. To find the energy of IR photons emitting at a wavelength of 1.03 × 10^-6 meters, we employ the relation between energy and wavelength of photons. The photon energy (E) is given by the Planck relation:

E = hν = hc / λ

where:

• h = Planck's constant = 6.626 × 10^-34 Joule·seconds
• c = speed of light = 3.00 × 10⁸ meters/second
• λ = wavelength = 1.03 × 10^-6 meters

Calculating the energy in Joules:

E = (6.626 × 10^-34) × (3.00 × 10⁸) / (1.03 × 10^-6) ≈ 1.93 × 10^-19 Joules

To convert Joules into electronvolts (eV), recall that 1 eV = 1.602 × 10^-19 Joules. Therefore:

E ≈ (1.93 × 10^-19) / (1.602 × 10^-19) ≈ 1.20 eV

Hence, the energy of the IR photons with wavelength 1.03 × 10^-6 meters is approximately 1.20 eV.

Next, computing the frequency of the gamma-ray photon with energy 0.779 MeV, we use the relation:

E = hν ⇒ ν = E / h

Convert energy from MeV to Joules: 1 eV = 1.602 × 10^-19 Joules, so

• 0.779 MeV = 0.779 × 10⁶ eV = 0.779 × 10⁶ × 1.602 × 10^-19 Joules ≈ 1.248 × 10^-13 Joules

Calculating frequency:

ν = (1.248 × 10^-13) / (6.626 × 10^-34) ≈ 1.88 × 10²⁰ Hz

This frequency aligns with the extremely high energy level of gamma-ray photons emitted during nuclear reactions.

De Broglie Wavelength of Electrons and Protons

The de Broglie wavelength (\(\lambda\)) embodies the wave-particle duality, describing particles such as electrons and protons (which exhibit wave-like properties). It is calculated using the relation:

\(\lambda = h / p\)

where p is the momentum of the particle.

Part 1: Electron with momentum 1.3 × 10^-24 kg·m/s

Calculate the de Broglie wavelength:

\(\lambda = 6.626 × 10^{-34} / 1.3 × 10^{-24} ≈ 5.09 × 10^{-10}\) meters

This wavelength (~0.509 nm) is within the range of electron wavelengths used in electron microscopy.

Part 2: Proton traveling at 7.0 × 10^-3 m/s

The momentum of the proton (mass = 1.67 × 10^-27 kg):

• p = m × v = 1.67 × 10^-27 kg × 7.0 × 10^-3 m/s = 1.169 × 10^-29 kg·m/s

Calculating the de Broglie wavelength:

\(\lambda = 6.626 × 10^{-34} / 1.169 × 10^{-29} ≈ 5.66 × 10^{-5}\) meters

This indicates that slow-moving protons have relatively large de Broglie wavelengths, often negligible in classical physics but significant in quantum regimes.

De Broglie Wavelengths in Hydrogen Atom: Orbit and Energy Transitions

In Bohr’s model, the radius of the electron’s orbit in a hydrogen atom is related to the de Broglie wavelength by the quantization condition:

\(\lambda = 2\pi r / n\)

Given:

• r = 1.905 nm = 1.905 × 10^-9 meters
• n = 1 (ground state for the initial orbit)

Calculating the de Broglie wavelength:

\(\lambda = 2\pi \times 1.905 × 10^{-9} / 1 ≈ 1.196 × 10^{-8}\) meters

This wavelength aligns with the quantized nature of the electron’s orbit, reinforcing de Broglie’s hypothesis that electrons exhibit wave-like behavior.

The momentum (\(p = m v\)) of the electron in this orbit is related via the classical relation:

p = h / \(\lambda\) ≈ 6.626 × 10^-34 / 1.196 × 10^-8 ≈ 5.54 × 10^-26 kg·m/s

Energy needed to ionize the electron from this level can be calculated via the energy difference between the n=1 level and free state, which from the Bohr model is:

E = -13.6 eV / n² = -13.6 eV / 1² = -13.6 eV

Converting to positive value (since energy must be absorbed): 13.6 eV.

The frequency of the photon needed to ionize the atom (from E = hν):

ν = E / h = 13.6 eV / 4.1357 × 10^-15 eV·s ≈ 3.29 × 10¹⁵ Hz

Emission and Absorption in Hydrogen: Transition Energies

The photon emitted when the electron transitions from n=2 to n=1 has an energy difference:

ΔE = E_n=1 - E_n=2 = -13.6 eV (for n=1) - (-13.6 eV / 4) (for n=2) = -13.6 + 3.4 = -10.2 eV

Absolute value: 10.2 eV, which is the energy of the emitted photon.

If the atom absorbs a photon to return from n=3 to n=2, the energy of the photon is:

ΔE = -13.6 eV / 4 - (-13.6 eV / 9) ≈ 1.89 eV

Converting into frequency:

ν = 1.89 eV / 4.1357 × 10^-15 ≈ 4.57 × 10¹⁴ Hz

Electronic Configuration of Aluminum and Electron Population in n=3 Level

Aluminum's atomic number is 13, which indicates it has 13 electrons. Its electronic distribution in the ground state fills the energy levels as follows:

• N=1: 2 electrons
• N=2: 8 electrons
• N=3: 3 electrons (the remaining electrons in the third level)

Thus, in the ground state, there are 3 electrons in the n=3 level, matching the configuration where the n=1 level contains 2 electrons and the n=2 level contains 8 electrons.

Nuclear Composition of Various Isotopes

• Iron-56: 26 protons, 30 neutrons
• Nickel-58: 28 protons, 30 neutrons
• Antimony-121: 51 protons, 70 neutrons
• Erbium-166: 68 protons, 98 neutrons
• Uranium-238: 92 protons, 146 neutrons

These values are derived from the atomic number (number of protons) and mass number (protons + neutrons) information typically associated with isotopes.

Radioactive Decay Reactions and Half-Life Calculations

The beta decay of lithium-8 proceeds as:

Li-8 → Be-8 + e^-

Similarly, alpha decay of astatine-201 involves emission of a helium nucleus:

At-201 → Bi-197 + He

The gamma decay of palladium-120 involves the emission of a gamma photon, leaving the nuclear configuration unchanged except for energy states.

To estimate the half-life from count rate data, use the decay formula:

N(t) = N₀ × (1/2)^{t / T_1/2}

Given initial counts per minute of 5200 and counts after six minutes of 1300, the decay constant (\(\lambda\)) can be found through:

5200 / 1300 = 4 = e^λt = e^6λ

Thus, λ = (1/6) × ln(4) ≈ 0.231 per minute

The half-life T_1/2 relates to λ via:

T_1/2 = ln(2) / λ ≈ 0.693 / 0.231 ≈ 3.0 minutes

Therefore, the isotope's half-life is approximately 3 minutes.

Regarding the contamination decay over days when the level is initially eight times permissible, the decay relation is:

N(t) = N₀ × (1/2)^{t / T_1/2}

Solving for t when N(t) = N₀ / 8:

(1/2)^{t / T_1/2} = 1 / 8 = (1/2)³

t / T_1/2 = 3 ⇒ t = 3 × T_1/2 = 3 × 4 days = 12 days

Thus, it takes approximately 12 days for the radioactivity to decay to acceptable levels.

Mass-Energy Equivalence in Nuclear Reactions

The energy released from matter-antimatter conversion is computed via Einstein’s mass-energy equivalence:

Energy = Δm × c²

Given:

• Δm = 5.4 grams = 0.0054 kg
• c = 3.00 × 10⁸ m/s

Calculating energy:

Energy = 0.0054 × (3.00 × 10⁸)² = 0.0054 × 9.00 × 10¹⁶ = 4.86 × 10¹⁴ Joules

This large energy release underscores the immense power in nuclear and matter-antimatter annihilation processes, exemplified by nuclear weapons testing.

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